What the figures prove — page 5
Logic
15 families
chains
36 kinds of claim · 23 placements
- person 1 is right on exactly half the assignments ×5
- the number of people is a whole number between 3 and 12 ×2
- a choice from each of the pairs is one of two to the power of their number
- after its last difference each sequence agrees with the representative for good
- and everyone after it is right
- between two and five pairs, each of two named things
- different pairs of whole numbers name different numbers a + b√2
- each element belongs to exactly one chain
- each rational point sits on the line through (1, f(1))
- every moved point lands strictly inside the open interval
- every point the map has to reach is reached
- f(x + y) = f(x) + f(y) on a + b√2
- no two moved points land on the same place
- on every one of the 2^8 assignments, everyone after the first is right
- the assembled map sends no two elements to the same place
- the function is not the line y = x
- the largest denominator is a whole number between 2 and 24
- the last wrong guess is at the last place the hats differ from the representative
- the left-to-right map sends different elements to different places
- the left-to-right shift is a whole number between 1 and 4
- the members of a pair are named apart exactly when an order is claimed
- the number of elements drawn from each side is a whole number between 5 and 14
- the number of sequence terms drawn is a whole number between 4 and 12
- the pairs either carry an order or they do not
- the plotted graph meets every cell of a six-by-six grid over the window
- the range of whole coefficients is a whole number between 2 and 6
- the right-to-left map sends different elements to different places
- the right-to-left shift is a whole number between 1 and 4
- the rule fits two to the power of the pairs it cannot tell apart
- the rule names exactly one of the rows exactly when the pairs carry an order
- the two maps are different, or there are no chains to see
- the value given to √2 is a number of modest size
- the view is one the family draws
- the window shows at least three chains
- under the parity strategy all are right together or all wrong together
- with √2 sent to √2 the function is the line y = x
connectives
8 kinds of claim · 7 placements
- between one and five bases are compared
- between two and eight connectives are tabulated
- no two of the sixteen have the same table
- Post's criterion and the closure agree about whether a connective is complete on its own
- the closure and Post's criterion agree about this basis
- the connective is one of the sixteen
- the number of columns is a whole number between 4 and 8
- there are sixteen binary connectives and no more
disc-model
62 kinds of claim · 19 placements
- the 1 axioms c > 0, …, c > 0 are satisfied by 1 and not by 0 ×6
- (a + b) + c = a + (b + c) holds on every sampled triple
- (a × b) × c = a × (b × c) holds on every sampled triple
- −1 appears at the first stage
- √2 appears at some stage drawn
- a + b = b + a holds on every sampled triple
- a × (b + c) = a × b + a × c holds on every sampled triple
- a × b = b × a holds on every sampled triple
- a < b and 0 < c imply a × c < b × c holds on every sampled triple
- a < b implies a + 1 ≤ b holds on every sampled triple
- a < b implies a + c < b + c holds on every sampled triple
- a ≠ 0 implies a = d + 1 for some d holds on every sampled triple
- a line drawn as parallel was checked and does not meet the base line inside the disc
- a triangle has three vertices
- between 4 and 9 terms
- between three and eight pieces, on a line long enough to hold them
- between two and four levels of galaxies
- each crossing point was computed and lies on both lines
- each dot is a root of its polynomial
- each lower truncation is below π and each upper one above
- each neighbourhood shows 2 to 4 steps either side
- each stage adds new elements
- every angle of the triangle is positive
- every drawn polynomial belongs to the structure
- every function from A to B is listed
- every geodesic arc meets the boundary circle at a right angle
- every ordinary number is one or the other
- every vertex is inside the disc
- exactly one of a < b, a = b, b < a holds on every sampled triple
- exactly one of the lines through the point never meets the base line
- heights from 2 to 7
- inside the model no function is onto
- more than one geodesic through the point misses the base line
- no drawn dot is a known transcendental
- nothing lies strictly between an element and its successor
- outside, the bijections number size factorial
- sets of two to four elements
- some element is neither even nor odd
- the angles of a triangle in this model add to less than a straight angle
- the defect is a positive area
- the drawn elements are in increasing order
- the even witness doubles back to the element
- the galaxies are distinct and in order
- the lower ones rise and the upper ones fall
- the model's functions stay not-onto after relabelling B
- the number of lines tried through the point is a whole number between 3 and 9
- the number of parallels drawn is a whole number between 1 and 9
- the odd witness gives the element
- the outside point is inside the disc
- the parity statement fails on sampled elements
- the point is off the base geodesic
- the view is one of triangle, euclid, compactness, polymodel, parity, axioms, galaxies, algebraic, skolemhull, relative, sentences, supremum
- the witness for "∀a∀b∀c ∃x x³ + ax² + bx + c = 0" checks
- the witness for "∀x (0 < x → ∃y y·y = x)" checks
- the witness for "∀x∀y (x < y → ∃z (x < z ∧ z < y))" checks
- the witness for "∃x (x·x = x + x ∧ ¬ x = 0 ∧ ¬ x = 1 + 1)" checks
- the witness for "∃x x·x = −1" checks
- the witness for "∃x x·x = 2" checks
- the witness for "∃x x·x·x = x + 1" checks
- three to nine elements of the structure
- two or three stages
- X and its whole neighbourhood exceed every ordinary number
ef-game
84 kinds of claim · 42 placements
- at 1 rounds the search agrees with the rule that says same length or both at least 1 ×3
- at depth 1 the words must be 1 letters before Duplicator survives ×3
- by the largest size drawn, Duplicator survives 2 rounds on every pair sampled ×2
- Duplicator survives 3 rounds on a cycle of 22 against two of 11 ×2
- the search agrees with the rule that says Duplicator survives 3 rounds from 5 points up ×2
- a degree-three point gives four middle points
- a star splits at the first round
- a vertex's neighbours induce different shapes in the two graphs, so they are not isomorphic
- an even number of edges is a coin toss at every size, as the flip bijection says it must be
- and at every point of the second
- and does not at the smallest
- and does not survive the three-pebble one
- and Duplicator survives at every length past it
- and every one of them swaps an even number of edges
- and it does grow
- and no pair survives three pebbles without surviving two — more pebbles never help Duplicator
- and once Duplicator survives at one length it survives at every longer one
- and so do two triangles, which is why refinement cannot separate them
- and so does the two-dimensional refinement, which separated the last pair at once
- and that object is a path
- and the group has order two
- and the shape is the same at every point of the first
- and the share rises with the size rather than wandering
- and the two graphs have the same number of edges
- and they have the same parameters
- at every depth there is a length past which the two words cannot be separated
- between fifty and four hundred graphs at each size
- between four and twelve pairs of graphs at each size
- between ten and eighty graphs are sampled at each size
- between three and six graph sizes, each of four to sixty points
- between three and six rounds shown
- between three and six sizes, each of four to forty-four points
- between three and six sizes, each of three to forty points
- between two and four depths
- between two and three round counts are tabulated
- both graphs are strongly regular
- Duplicator survives the two-pebble bijective game
- every point of both graphs has two neighbours
- exactly four of the eight sign patterns are realised
- exactly half of the 64 graphs on 4 points have an even number of edges
- large ones almost never do
- no odd pattern is realised, which is the whole of what the gadget enforces
- one graph is connected and the other is not, so a sentence separating them would express connectedness
- one or two round counts, each of at most three
- one-dimensional refinement gives both graphs the same colours and cannot tell them apart
- one-dimensional refinement gives them the same colours
- out of the eight subsets of its three edges
- six neighbours each
- sixteen points each
- small graphs usually fail the property
- so the count grows like a logarithm rather than like the length
- some pair survives two pebbles and not three, which is the point of a third variable
- the cycle stays one colour at every round, since every point looks alike
- the deeper game needs a larger graph before Duplicator always survives it
- the depth is between one and three
- the first-order property "every pair has a common neighbour" has settled on 1 by the largest size drawn
- the first-order property "some point joined to nothing" has settled on 0 by the largest size drawn
- the first-order property "three points all joined" has settled on 1 by the largest size drawn
- the game agrees with the neighbourhood argument about who wins
- the game is played between two and five rounds deep
- the game is played to at most three rounds, which is as far as a search finishes
- the game runs between one and four rounds
- the measured share agrees with the count of expected failures
- the number of colours never falls
- the number of rounds Spoiler needs is computed
- the other two monoids are aperiodic, so both languages are first-order definable
- the parity language's monoid contains a non-trivial group, so no first-order sentence says it
- the rounds needed against a chain one longer follow the logarithm of its length
- the sizes are listed in increasing order
- the sweep runs over between three and eight cycle lengths
- the sweep runs to between five and nine
- the table runs to between four and nine
- the transcript ends the way the exhaustive search says the game ends
- the two cycles are of different lengths
- the two cycles have three points each
- the two graphs have the same number of points and the same number of edges
- the two neighbourhoods are the same object, computed from the cycles rather than named
- the two small cycles run to between three and thirteen points
- the view is one the family draws
- two chains of between two and nine elements
- two cycles of between five and thirty points
- two-dimensional refinement does tell them apart
- words up to between four and six letters
- words up to between six and twelve letters
infinite
32 kinds of claim · 22 placements
- a set of 0 has more subsets than members ×13
- height 2 holds finitely many polynomials, and the enumeration produced them ×5
- −1 is among the roots the enumeration found
- √2 is among the roots the enumeration found
- 1 is among the roots the enumeration found
- a half is among the roots the enumeration found
- a larger height admits more polynomials
- all 512 listings of 3 subsets of a set of 3 were tried
- and the second
- and yet the two images are far apart, so the weave is not continuous
- by a factor of more than a thousand at this many places
- every fraction the grid reaches is on the list
- every kept cell got exactly one place in the list
- no fraction is listed twice
- the complemented diagonal is on no row of the listing
- the digit strings are as long as the drawing says
- the first number's digits are between four and ten decimal places
- the golden ratio is among the roots the enumeration found
- the largest height enumerated is a whole number between 3 and 6
- the largest sum of numerator and denominator drawn is a whole number between 3 and 9
- the number of places drawn is a whole number between 4 and 10
- the number of rungs drawn is a whole number between 3 and 5
- the places run from one with no gaps
- the second number has the same number of places
- the size at which every listing is tried is a whole number between 2 and 3
- the two expansions differ by less than one unit in the last place drawn
- the view is one the family draws
- the walk's count and the sum of totients agree, having been computed independently
- the window holds enough of them to draw
- the window on the line is between one and twelve wide
- the woven number has one place from each, alternately
- unweaving recovers the first number
karnaugh
12 kinds of claim · 9 placements
- consecutive entries of the Gray code differ in exactly one bit
- every cell of a prime implicant is one the function makes true
- every leaf of the formula is a variable
- every opening bracket in the formula is closed
- squares next to each other on the map differ in one variable
- the covering rectangles and the formula agree on every assignment
- the formula and its disjunctive normal form agree on every row
- the formula has between one and five variables
- the formula is a non-empty string
- the map is drawn for three or four variables
- the table has one row per assignment
- the whole formula is consumed by the parser
kripke
72 kinds of claim · 40 placements
- □□p → □p is valid exactly on the dense frames
- □p → □□p is valid exactly on the transitive frames
- □p → ◇p is valid exactly on the serial frames
- □p → p is valid exactly on the reflexive frames
- □p at a world is p at every world it can reach
- ◇□p → □◇p is valid exactly on the confluent frames
- ◇p → □◇p is valid exactly on the euclidean frames
- ◇p at a world is p at some world it can reach
- 4 is valid on a frame exactly when the frame is transitive
- 5 is valid on a frame exactly when the frame is euclidean
- a formula separating the other pair exists among those searched
- a transitive model filters to a transitive quotient here, checked rather than assumed
- a world and a valuation were found at which the axiom fails
- a world that reaches nowhere makes every necessity true and every possibility false
- and fewer worlds than the model it came from
- and it is deeper than anything the closure could ask
- and no frame with any worlds in it validates both Löb's axiom and □p → p
- and on every one of them ¬□⊥ is a fixed point of ¬□p
- and some class of the quotient can — the collapse invents a way back that the model has none of
- and the other two starting worlds are not
- and two frames agreeing on all of them disagree about the axiom
- and validates what is valid
- B is valid on a frame exactly when the frame is symmetric
- between one and five axioms are tabulated
- between two and six frames are compared
- D is valid on a frame exactly when the frame is serial
- excluded middle at a stage is exactly p there or ¬p there
- formulas up to depth two to four are compared
- Löb's axiom is valid exactly on the transitive frames with no cycles
- no condition in the catalogue matches McKinsey's axiom
- no stage forces both p and its negation
- no world of the model can be returned to
- on a frame with the property the axiom holds everywhere, under every valuation
- on that frame every deeper modality agrees with the first, not only the second
- one of the two frames collapses the tower and the other does not
- p → □◇p is valid exactly on the symmetric frames
- so every frame of the logic is transitive, which Löb's axiom implies
- some depth of looking ahead separates the chain from its quotient
- some frame this family knows has the property
- some frames validate it
- some stage forces neither p nor its negation, so p ∨ ¬p fails there
- T is valid on a frame exactly when the frame is reflexive
- the axiom is one of the five
- the bisimilar worlds agree on every formula up to the drawn depth
- the chain runs between eight and twenty-four worlds
- the deeper closure is asked for or not
- the frame is one this figure knows
- the gallery's verdict for a branch, both ends dead is the sweep's
- the gallery's verdict for a single dead end is the sweep's
- the gallery's verdict for a transitive chain is the sweep's
- the gallery's verdict for a two-cycle is the sweep's
- the gallery's verdict for a world seeing itself is the sweep's
- the gallery's verdict for one step and stop is the sweep's
- the model is the chain or the strict order on its worlds
- the model refutes □p → p, which is not valid in this logic
- the quotient has at most one world for each way of answering the closure
- the sweep runs over frames on three or four worlds
- the tower is followed to between two and four boxes
- the truth lemma holds for □□p: it is true at a world exactly when it is true at that world's class
- the truth lemma holds for □p: it is true at a world exactly when it is true at that world's class
- the truth lemma holds for p: it is true at a world exactly when it is true at that world's class
- the truth lemma holds: □□p is true exactly where it is accepted
- the truth lemma holds: □p is true exactly where it is accepted
- the truth lemma holds: p is true exactly where it is accepted
- the two starting worlds are related by the bisimulation
- the view is one the family draws
- this frame is not reflexive, so the axiom has a chance of failing
- this frame is not symmetric, so the axiom has a chance of failing
- this frame is not transitive, so the axiom has a chance of failing
- two frames this family knows
- what is established at a stage stays established later
- worlds share a class exactly when no formula of the closure tells them apart
lattice
55 kinds of claim · 42 placements
- the size of the underlying set is a whole number between 2 and 4 ×3
- a cyclic order has exactly n arcs of length k
- an element's share of the maximal chains is one over its layer's size
- and a star is a family of that size
- and every chain meets the middle layer exactly once
- and every one of them is a whole layer
- and every two of its members really do meet
- and everything sits below something maximal
- and some antichain's shares add to exactly one
- and takes one step of size at each rank between its ends
- at most k of the n arcs pairwise meet, and k of them do
- each chain is symmetric about the middle of the cube
- each chain runs upward in the order
- each named target is a subset index is a whole number between 0 and 7
- each step adds exactly one element
- each subset is joined to the ones one element larger
- every chain has an upper bound inside the order
- every element of the order is on some chain
- every relation of the order is respected by every extension
- every subset is on exactly one chain
- exactly n + 1 antichains use the chains up exactly
- no antichain's shares add to more than one
- no chain meets the antichain twice
- no element is on two chains
- no element is sent to the built set
- no subset is on two chains
- no two elements of one layer are comparable
- so there is something maximal
- some incomparable pair splits the extensions between a third and two thirds
- the built set disagrees with f(k) about whether k belongs
- the chain count is the elements less the matched pairs
- the extensions enumerated and the extensions counted by peeling agree
- the fewest chains covering the order is the size of the largest antichain
- the largest antichain is at least as large as the widest layer
- the largest family in which every two sets meet is as large as a star
- the map names one subset per element
- the number has between four and sixteen divisors
- the number has between four and sixteen proper divisors
- the number of chains is the size of the middle layer
- the number of subsets is two to the power of the set's size
- the number whose divisors are ordered is a whole number between 4 and 210
- the number whose divisors are ordered is between 4 and 210
- the number whose proper divisors are ordered is a whole number between 6 and 210
- the order does not already decide everything
- the order has an incomparable pair to ask about
- the order is one this family draws
- the order is one this mode draws
- the order is small enough to search every collection of its elements
- the order is small enough to search every subset of
- the sets of the right size were all generated
- the size of the sets in the family is a whole number between 2 and 4
- the theorem needs the sets to be small enough that two can miss each other
- the two ends of an edge differ by exactly one element
- the view is one the family draws
- there are more subsets than elements
opens
41 kinds of claim · 15 placements
- ¬¬¬p → ¬p holds in every algebra searched
- ¬p → q ∨ r ⟹ (¬p → q) ∨ (¬p → r) holds in every algebra searched
- a set is contained in its double negation
- and here it is strictly contained — the punched-out points come back
- and it keeps rising, so the classes are not exhausted by short formulas
- and so does double-negation elimination
- and the value really is not the top
- between one and four points are punched out
- every element sits below its double negation
- every punched-out point is inside the set
- every smaller algebra validates (p → q) ∨ (q → p)
- every smaller algebra validates ¬¬p → p
- every smaller algebra validates ¬p ∨ ¬¬p
- every smaller algebra validates p ∨ ¬p
- excluded middle fails somewhere in this algebra
- formulas of up to 3 to 6 nestings
- more than four formulas in one variable are distinguished
- no open set larger than the implication satisfies the condition
- some axiom holds in one algebra and fails in another
- the algebra has at least two elements
- the count never falls as the formulas grow
- the formulas are among em, wem, dnn, gd, kp, triple
- the implication really does meet the antecedent inside the consequent
- the implication really is the largest element whose meet lies below
- the negation of an open set is open
- the orders are among point, chain2, chain3, anti2, v, lambda, diamond
- the refuting algebra has a valuation making (p → q) ∨ (q → p) fall short
- the refuting algebra has a valuation making ¬¬p → p fall short
- the refuting algebra has a valuation making ¬p ∨ ¬¬p fall short
- the refuting algebra has a valuation making p ∨ ¬p fall short
- the search has several algebras in it
- the search runs over orders on up to 2 to 4 points
- the search separates the formulas it refutes from the ones it cannot
- the separating algebras run to 2 to 4 points
- the set sits strictly inside the ambient line
- the set together with its negation does not fill the line
- the union has as many pieces as the two sets between them
- the view is one the family draws
- three negations are the same as one
- three negations collapse in every Heyting algebra
- three negations collapse to one
ordinal
45 kinds of claim · 35 placements
- the number of ticks per block is a whole number between 4 and 14 ×2
- a fundamental sequence belongs to a limit ordinal
- a larger ordinal grows at least as fast at every argument
- a limit mark is where the marks before it pile up
- a successor is one step past the ordinal below it
- an ordinal expression is built from whole numbers, w, +, * and ^
- and so do the two multiplications on the other side
- and the terms strictly increase
- at least one entry needs an exponent that is itself a power
- at least one of the ordinals drawn is a limit
- between one and five order types are drawn
- between one and four pairs of expressions are compared
- between three and six rates are drawn
- between two and six ordinals are drawn
- each number's ordinal is strictly larger than the one before it
- each part of the expression is a number, w, or a bracket
- each rate is increasing in its argument
- every limit mark has marks before it
- every limit mark has marks piling up before it
- every term of the sequence is below the ordinal it approaches
- F₀ adds one
- F₁ doubles its argument
- F₂ multiplies by two to the power of its argument
- nothing sits between the pile-up and the limit it accumulates at
- ordinal addition is associative on the drawn values
- the arrow clears both the last term and the supremum it points at
- the base the numbers are written in is a whole number between 2 and 4
- the brackets in the expression close
- the drawn pair is an ordinal of the form ω·k + m with k between 1 and 3
- the exponents fall and the coefficients are positive
- the hereditary form reads back as the number it came from
- the index into a fundamental sequence is a positive whole number
- the largest argument in the table is a whole number between 3 and 5
- the largest number in the table is a whole number between 6 and 34
- the marks along each line strictly increase
- the number of steps shown is a whole number between 3 and 12
- the number of terms shown is a whole number between 3 and 7
- the ordinal beside each term is strictly smaller than the one before it
- the ordinal view is one of goodstein, arith, cnf, fund, growth
- the pair drawn on the lines is one of the pairs in the table, or null for none
- the starting value is a whole number between 2 and 6
- the two additions agree
- the two multiplications by a whole number agree
- the whole expression was read
- zero has nothing below it to approach
relation-grid
46 kinds of claim · 32 placements
- column 2 ×59
- column 2 is empty exactly when 2 is prime ×59
- slice a = 1: a solution exists exactly when the eliminated formula says so ×2
- the formula for ∀x ∃y at n = 2 ×2
- ∀x ∀y R(x, y) holds for R exactly when its dual fails for the complement
- ∀x ∀y R(x, y) implies ∃x ∀y R(x, y) with nothing between
- ∀x ∀y R(x, y) implies ∃y ∀x R(x, y) with nothing between
- ∀x ∃y R(x, y) holds for R exactly when its dual fails for the complement
- ∀x ∃y R(x, y) implies ∃x ∃y R(x, y) with nothing between
- ∀y ∃x R(x, y) holds for R exactly when its dual fails for the complement
- ∀y ∃x R(x, y) implies ∃x ∃y R(x, y) with nothing between
- ∃x ∀y R(x, y) holds for R exactly when its dual fails for the complement
- ∃x ∀y R(x, y) implies ∀y ∃x R(x, y) with nothing between
- ∃x ∃y R(x, y) holds for R exactly when its dual fails for the complement
- ∃y ∀x R(x, y) holds for R exactly when its dual fails for the complement
- ∃y ∀x R(x, y) implies ∀x ∃y R(x, y) with nothing between
- 7³ = 343 relations have a mark in every row
- a column holds a mark exactly when its values of a satisfy |a| ≥ 2
- a full column forces every row to carry a mark, so one reading implies the other
- almost every cell is checked against the formula
- and 512 − 343 = 169 have a full row
- and for ∃x ∀y
- at 16 points almost every relation has a mark in every row
- every x makes it positive exactly when a² − 4b < 0
- no drawn row agrees with the built row everywhere it has been checked
- no row of the table is the Russell row
- slice a = -1: a solution exists exactly when the eliminated formula says so
- some relation satisfies ∀x ∃y R(x, y) but not ∀y ∃x R(x, y)
- some relation satisfies ∀x ∃y R(x, y) but not ∃y ∀x R(x, y)
- some relation satisfies ∀y ∃x R(x, y) but not ∀x ∃y R(x, y)
- some relation satisfies ∃x ∀y R(x, y) but not ∀x ∃y R(x, y)
- some x makes it zero exactly when a² − 4b ≥ 0
- the built row differs from row n in place n
- the diagram has six arrows
- the drawn window is at least as wide as it is tall
- the labelling is one this figure knows
- the largest number is a whole number between 20 and 80
- the number of columns is a whole number between 2 and 14
- the number of drawn columns is a whole number between 3 and 16
- the number of drawn rows is a whole number between 3 and 12
- the number of rows is a whole number between 2 and 14
- the number of sets in the table is a whole number between 3 and 10
- the relation is one this figure knows how to draw
- the Russell row disagrees with row k at column k
- the search agrees with 256b³ ≤ 27a⁴
- the share with every row marked eventually only rises
syllogism
8 kinds of claim · 7 placements
- a form valid without existential import is valid with it
- every letter in the set expression names a drawn curve
- fifteen forms are valid with no assumption that anything exists
- nine more become valid once every term is assumed non-empty
- the form is three mood letters and a figure number, like AAA-1
- there are 256 syllogistic forms — four figures and sixty-four moods
- twenty-four forms are valid in the traditional list
- two circles realise all four patterns
tree
147 kinds of claim · 54 placements
- the chain on 2 variables is refuted by a search of 5 nodes ×6
- the clauses for 2 pigeons in 1 holes ×6
- at 1 instances there is still a model, so the search has not finished ×5
- the assumption h2 is open where it is discharged ×3
- at 1 instances there is still a model ×2
- the assumption h1 is discharged somewhere below it ×2
- the variable h1 is bound where it is used ×2
- a conjunction is introduced from two derivations
- a conjunction is taken apart from one derivation
- a disjunction is introduced from one derivation
- a disjunction is used by considering both halves
- A implies B on every assignment
- A implies the interpolant
- a negation is introduced from a derivation and its denial
- a variable shares a component with its negation exactly when no assignment works
- an assumption has nothing above it
- an implication is introduced from one derivation
- an introduced implication is an implication
- and both cases reach the same conclusion
- and every assumption it made has been discharged
- and it is discharged as (p -> q) & (q -> r)
- and it is discharged as p
- and it is discharged as p -> p
- and it is discharged as p & (q | r)
- and it is discharged as q
- and it is discharged as r
- and it is the conjunction of what they derived
- and its antecedent is the discharged assumption
- and its argument has the antecedent's type
- and its consequent is what was derived
- and once it closes it stays closed
- and the count of models is positive at every stage drawn
- and the drop is sharper for more variables
- and the formula it proves is true in every row of its table
- and the goal follows from the lemma
- and the goal is true in every row of its table
- and the interpolant implies B
- and the lemma the other proof uses is not a subformula of the goal
- and the named half is the conclusion
- and the named half is what was derived
- and the negated assumption is the conclusion
- and the second is its antecedent
- and the sweep contains both kinds
- between two and seven stages
- both cases have the same type
- cases are taken on a disjunction
- each arrow along the chain resolves with the clause built so far
- each extra hole multiplies the search by more than the last one did
- every drawn step is the resolvent of the two clauses above it
- every formula in the cut-free proof is a subformula of the goal, or its negation
- every formula of the derivation with no detour is a subformula of the goal
- every label is falsified by the assignments on its own path
- every leaf of the formula is a variable
- every level has a node that reaches the bottom
- every literal names a variable of the clause set
- every literal names a variable of the set
- every opening bracket in the formula is closed
- every order of the six variables
- it closes exactly when the universal has been used 3 times
- modus ponens takes two derivations
- no arrow leads from a true literal to a false one
- no fixed order beats the best search that chooses afresh at each node
- no label is a tautology
- no node has more children than the branching factor
- no variable shares a component with its negation
- only a disjunction is taken apart by cases
- only a pair is projected
- only a term of implication type is applied
- removing a detour leaves the proved formula unchanged
- resolution and the truth table agree on every one of the sets
- resolution reaches the empty clause exactly when no assignment satisfies the set
- resolution refutes A together with the negation of B
- so the conclusion is its consequent
- so the count of unsatisfiable sets and the count of refuted sets are one number
- the assignment satisfies every clause
- the assumption f is discharged somewhere below it
- the assumption x is discharged somewhere below it
- the assumption y is discharged somewhere below it
- the branching factor is a whole number between 2 and 3
- the case assumption h2 is open
- the chain from x to y ends in the clause ¬x ∨ y
- the clause set is one the arrow views know
- the clause set is one the family knows
- the conclusion of →I is an implication
- the conclusion of ¬I is a negation
- the conclusion of ∨I is a disjunction
- the cut-free proof closes, so the formula is a theorem
- the depth drawn is a whole number between 3 and 7
- the derivation ends at its goal
- the derivation ends at the stated goal
- the derivation had at least one detour to remove
- the derivation is one of contraposition, transitivity, distribute
- the derivation is one of contraposition, transitivity, distribute, identity, swap
- the derivation is one of identity, swap
- the derivation uses only implication, conjunction and disjunction, where every step has a term
- the derivation's term has the goal as its type
- the expansion does close, at some stage this figure reaches
- the first derivation gives a disjunction
- the first is an implication
- the formula and its disjunctive normal form agree on every row
- the formula has between one and five variables
- the formula is a non-empty string
- the interpolant uses only atoms the two formulas share
- the last term contains no detour
- the lemma is itself a theorem, so it may be cut in
- the most holes is a whole number between 3 and 6
- the path passes through one node per level
- the propagating search only runs on unsatisfiable sets
- the pruning parameter is a whole number between 2 and 9
- the question is one of reaches, never
- the reduction finishes within a dozen steps
- the root's label is the empty clause
- the search only runs on unsatisfiable sets
- the set has a variable in its negation's component
- the sets with a variable in its negation's component are the unsatisfiable ones
- the short form agrees with the extracted interpolant on every shared assignment
- the smallest unsatisfiable set of two-literal clauses needs four of them
- the strongest interpolant implies the one read off the refutation
- the sweep is over three variables
- the table has one row per assignment
- the tableau and the truth table reach the same verdict
- the tableau stays small enough to draw
- the tableau tests validity or satisfiability
- the term's type at this node is the formula the derivation has there
- the tree drawn actually reaches the bottom row
- the trials per point is a whole number between 20 and 400
- the truth table, resolution and the components agree on every set
- the two derivations are a formula and its denial
- the two formulas share at least one atom
- the two formulas use at most six atoms between them
- the two unit clauses resolve to the empty clause
- the variable f is bound where it is used
- the variable h is bound where it is used
- the variable k is bound where it is used
- the variable x is bound where it is used
- the variable y is bound where it is used
- the view is one the family draws
- the walk always has a surviving child to step to
- the whole formula is consumed by the parser
- the whole term is closed, because every assumption was discharged
- there are twelve two-literal clauses on three variables
- this view draws a refutation, so the set has to be unsatisfiable
- two to four numbers of variables from 10 to 2000
- which derived a conjunction
- which implies the weakest
- with few clauses nearly every random set can be satisfied
- with half as many clauses again as variables, the largest sets almost never can
truth-table
27 kinds of claim · 22 placements
- every corner meets one edge per variable, each edge counted once
- every leaf of the formula is a variable
- every opening bracket in the formula is closed
- every set of more than half the corners has a corner of degree at least √n
- majority of p, q, r is a weighted vote
- no function has degree above its sensitivity squared
- on majority the rule stops making mistakes
- on parity it never does
- p, or both q and r is a weighted vote
- parity is sensitive to every letter everywhere
- parity: an odd number of p, q, r is not a weighted vote
- the counts are 4, 14, 104 and 1,882
- the cube is drawn for two, three or four variables
- the formula and its disjunctive normal form agree on every row
- the formula has between one and five variables
- the formula is a non-empty string
- the function is one this figure knows
- the highlighted rows are the true ones, the false ones, or neither
- the matrix squared is n times the identity
- the nonzero entries are the cube's edges
- the search is over the 3-cube or the 4-cube
- the signed matrix is drawn for one to four letters
- the table has one row per assignment
- the two ends of an edge differ in exactly one variable
- the view is one of cube, threshold, thresholdcount, perceptron, sensitivity, huang, signed, degree
- the weights and the cut give the function
- the whole formula is consumed by the parser
venn
11 kinds of claim · 9 placements
- 3 circles realise every one of the 8 patterns ×2
- 4 circles cannot realise all 16 patterns ×2
- and each of the sixteen is a single connected region
- every leaf of the formula is a variable
- every letter in the set expression names a drawn curve
- no pattern occupies more pieces than the arrangement has
- the formula is a non-empty string
- the four ellipses realise all sixteen patterns
- the number of sets is a whole number between 2 and 5
- the pieces the circles cut the plane into match Euler's count
- the whole formula is consumed by the parser
Computation
11 families
check-digit
8 kinds of claim · 2 placements
- a scheme that weights every position the same catches no transposition
- and every single-digit error too
- descending weights over a prime modulus catch every transposition
- every single-digit error is caught exactly when the modulus is at least ten and coprime to every weight
- the error kinds are single, transpose or twin
- the number is between four and fourteen digits
- the number is between six and fourteen digits
- the scheme is one of sum-9, sum-10, flat-11, weighted-10, isbn-11
construct
151 kinds of claim · 63 placements
- 2cos(2π/11) is a root of it ×3
- the circle about P1 has a centre and a radius ×2
- the intersection named P1 is the one meant ×2
- the objects of step P1 meet ×2
- a collapsing compass cannot carry a length from one place to another
- a rational is a whole numerator over a non-zero whole denominator
- a right angle can be trisected
- a root found by the theorem really is a root of the cubic
- a straight angle can be trisected
- a zero angle can be trisected
- AL equals BC, as Euclid's second proposition promises
- and above it
- and at least one is out of reach of both
- and exactly half way from the other end too
- and is not half way along after it
- and it fails when the given point is not the midpoint, so the midpoint is what is used
- and it is a different line from the segment
- and it is one radius beyond B
- and it lies on the segment
- and it sits over the midpoint
- and on the segment
- and that crossing and the marked point are both on the circle
- and the diameter really is twice the radius
- and their images are not equally spaced
- at least two polygons separate the two instrument sets
- between three and eight polygons, each with 3 to 24 sides
- both crossings exist
- CB′ equals AB: reflecting in the line XY carries A to C and B to B′
- copying a length with a rigid compass costs four operations
- corner A is at 0°
- corner B is at 60°
- corner C is at 120°
- corner D is at 180°
- corner E is at 240°
- corner F is at 300°
- cos(θ/3) is a root of the triple-angle cubic
- divisors are taken of a positive whole number
- dropping the ruler makes the midpoint dearer
- each step lands short of the far end
- each step meets the circle
- each step of the compass round the circle meets it
- Euclid's route to copying a length is dearer than a reflection with the same compass
- every coefficient of the minimal polynomial is a whole number
- every image lands on the second line
- every point of the first round is where two drawn objects cross
- everything the classical pair reaches, a conic reaches too
- Gauss's condition says the degree is a power of two
- Gauss's criterion and the degree test agree about every polygon
- Gauss's criterion and the degree test agree about the tripled polygon
- nineteen of the polygons up to sixty can be drawn
- one round puts two points off the line
- six steps of the radius come back exactly to the start
- sixty degrees cannot be trisected
- some angle in the table the two instruments cannot cut, and the marked one can
- the angle comes from a polygon that can be drawn
- the angle is between 12 and 174 degrees
- the angle the placed straightedge makes is a third of the original
- the apex is near the segment
- the apex sees the diameter at a right angle
- the centre is equidistant from the ends of the diameter
- the circle about A has a centre and a radius
- the circle about B has a centre and a radius
- the circle about C has a centre and a radius
- the circle about D has a centre and a radius
- the circle about E has a centre and a radius
- the circle about F has a centre and a radius
- the circle about O has a centre and a radius
- the circle about X has a centre and a radius
- the circle about Y has a centre and a radius
- the circle has a sensible radius
- the constructed line is parallel to the diameter
- the construction is one of those scored
- the cross ratio is the same on both lines
- the crossing is equidistant from the two ends of the remaining piece
- the degree of the cosine is half of n minus one
- the degree's smoothness and the Pierpont condition agree
- the denominator the cosines are taken over is a whole number between 2 and 16
- the diameter is tilted between 5 and 175 degrees
- the fixed opening is fixed
- the free point sits inside the segment PM
- the intersection named B is the one meant
- the intersection named B' is the one meant
- the intersection named C is the one meant
- the intersection named D is the one meant
- the intersection named E is the one meant
- the intersection named F is the one meant
- the intersection named G is the one meant
- the intersection named L is the one meant
- the intersection named M is the one meant
- the intersection named P is the one meant
- the intersection named Q is the one meant
- the intersection named X is the one meant
- the intersection named Y is the one meant
- the inverting circle meets the circle of radius AB
- the largest polygon considered is a whole number between 12 and 40
- the largest polygon tested is a whole number between 12 and 300
- the marked segment between the line and the circle is the radius
- the marked straightedge cuts every angle in the table
- the marks start equally spaced
- the middle mark is half way along before the projection
- the new height is √3/2
- the nine-gon is not — 3 is a Fermat prime but 9 repeats it
- the number being factorised is a whole number between 1 and 1000000
- the number of columns in the grid is a whole number between 6 and 20
- the number of compass steps is a whole number between 2 and 12
- the number of construction rounds is a whole number between 1 and 2
- the number of equally spaced marks is a whole number between 3 and 9
- the number of marks is odd so one of them is the midpoint
- the number whose square root is constructed is between 0 and 12
- the objects of step B meet
- the objects of step B' meet
- the objects of step C meet
- the objects of step D meet
- the objects of step E meet
- the objects of step F meet
- the objects of step G meet
- the objects of step L meet
- the objects of step M meet
- the objects of step P meet
- the objects of step Q meet
- the objects of step X meet
- the objects of step Y meet
- the perpendicular at the join has height √n
- the Pierpont condition and the degree's smoothness agree
- the point found is exactly half way along
- the point found is halfway along
- the polygon has a prime number of sides, so the degree is (n − 1)/2
- the polygon is a whole number between 5 and 23
- the polynomial has no rational root, so it does not factor off a linear piece
- the polynomial has the degree the totient predicts
- the second line is tilted between 0.15 and 0.75
- the second round is shown or not
- the segment cut off is longer than the radius far out and shorter close in
- the segment is between 0.4 and 6 units long
- the segment is between half a unit and two units long
- the segment really is longer than the opening can span in one go
- the seven-gon is not
- the seventeen-gon is constructible
- the sliding point and the centre are the same distance from the near crossing
- the totient of n above two is even
- the two arcs about D and E meet
- the two arcs of the fixed opening meet
- the two crossings sit at the same height, so the line is parallel
- the view is one of root, trisect, polygons, compass, hexagon, straightedge, neusis, reach, parallel, poncelet, rusty, conics, hendecagon, lemoine, lemoinetable
- the walk was built
- three steps of the radius reach the point twice as far away
- three steps reach the far side, which is the doubling
- twenty-four of the polygons up to a hundred can be drawn
- two points and one round of drawing give four more
- what is left is short enough for the fixed opening to span
- which makes it that piece's midpoint
cube-code
41 kinds of claim · 26 placements
- step 1 changes exactly one place ×7
- the dimension of the cube is a whole number between 2 and 4 ×3
- a code is more than one word and fewer than all of them
- a tour with changes 4, 4, 4, 4 exists and the search found it
- a tour with changes 8, 6, 6, 6, 6 exists and the search found it
- and its worst single step changes all 4 places at once
- and none twice
- and the changes add up to the number of steps
- and the last corner is one step from the first, so the walk closes
- and visits none of them twice
- counting up in the ordinary way changes more places than that
- each cycle was met once in each direction
- each step changes one place
- each step, the last back to the first included, swaps one element out and one in
- every codeword has the length the cube has dimensions
- every corner meets one edge per coordinate
- every place changes an even number of times
- every step of the code changes exactly one place
- every subset appears
- every walk is enumerated for at most the 4-cube
- every word within 1 of a codeword decoded back to it
- the 3-cube carries 6
- the 3-cube carries exactly 6 such walks
- the 4-cube carries 1,344 tours
- the 4-cube carries exactly 1,344
- the balls partition the cube exactly, so the code is perfect
- the code is a word list or one of repetition, parity, none, even2
- the cycle visits every word of the two levels once
- the drawn walks are the first of the ones counted
- the half-length k is a whole number between 1 and 4
- the length of the words is a whole number between 2 and 4
- the levels hold every word
- the number of places is a whole number between 3 and 11
- the radius of the balls drawn is a whole number between 0 and 2
- the search found a cycle through the middle two levels
- the size of the set is a whole number between 3 and 7
- the size of the subsets is a whole number between 1 and 5
- the tour is one of reflected, balanced
- the tour visits every corner once
- the view is one the family draws
- the walk visits all 8 corners
degree
68 kinds of claim · 44 placements
- φ^0 is 1 + 0φ ×10
- φ^1 has norm ±1 ×9
- the highest power is a whole number between 5 and 12 ×2
- √2 + ∛3 has degree six
- ∛2 × ∛2 lands on a multiple of ∛2²
- ∛2 × ∛2² lands on a multiple of 1
- ∛2 × 1 lands on a multiple of ∛2
- ∛2² × ∛2 lands on a multiple of 1
- ∛2² × ∛2² lands on a multiple of ∛2
- ∛2² × 1 lands on a multiple of ∛2²
- 1 × ∛2 lands on a multiple of ∛2
- 1 × ∛2² lands on a multiple of ∛2²
- 1 × 1 lands on a multiple of 1
- a bigger subgroup names a smaller field, never the other way round
- a rational is a whole numerator over a non-zero whole denominator
- a subgroup of order two fixes exactly one root
- and grow without bound
- and the product's at √2 · φ
- and the sign alternates
- both are monic
- divisors are taken of a positive whole number
- each ±√2 + ω^j∛3 is a root of the polynomial
- each number under a root is a whole number between 2 and 40
- each number under a root is square-free, or the step is not a real step
- each trace step divides exactly, as it must for an integer matrix
- every coefficient is a whole number under 400
- every degree divides the degree of the field, six
- every product of basis elements is a multiple of a basis element
- every rational root found by sweeping is one the theorem listed
- no integer polynomial in the searched range vanishes at π
- no power of two is a multiple of three
- no whole number the rational-root theorem allows is a root of x³ − m
- some combination has a smaller degree than the bound
- the cell of ℤ[√5] has area 2√5, twice as large
- the cell of ℤ[φ] has area √5
- the characteristic polynomial of the sum's matrix vanishes at √2 + φ
- the closest miss is a genuine miss
- the coefficient bound is a whole number between 2 and 8
- the coefficient range is a whole number between 2 and 6
- the control polynomial really vanishes at its algebraic number
- the cube root really cubes to it
- the degree is at most the product of the two degrees
- the degree of a tower of square roots is a power of two
- the degree of the tower is the size of its basis
- the denominators of the non-integer's powers never shrink
- the earlier powers are independent, so the solve is unique
- the exact value and the decimal one agree
- the grid half-width is a whole number between 3 and 8
- the group of six permutations has exactly six subgroups
- the largest degree searched is a whole number between 1 and 4
- the leading coefficient is not zero
- the monic test agrees with a and b having the same parity
- the number being factorised is a whole number between 1 and 1000000
- the number being searched for is one of pi, sqrt2, cbrt2, phi
- the number whose cube root is taken is between 2 and 30
- the polynomial found vanishes at √2 + ∛3
- the polynomial has degree between one and four
- the polynomial is named for the caption
- the polynomial vanishes at the number
- the powers of two are checked out to between 6 and 20 steps
- the product is the coefficient times the basis element it lands on
- the same search does find a polynomial for an algebraic number
- the searched number is a root of one of them
- the size of the subgroup times the degree of its field is the degree of the whole
- the tower has one basis element per subset of its roots
- the tower is built from one, two or three square roots
- the view is one the family draws
- φ^k lies in ℤ[√5] exactly when k is a multiple of 3
dissect
142 kinds of claim · 32 placements
- the point 0.041, 0.068, 0.026 lies in exactly one of the six pieces ×729
- the left piece keeps its 1th edge at 0° ×21
- the right piece keeps its 1th edge at 0° ×21
- the left piece still touches its pin at 0° ×7
- the right piece still touches its pin at 0° ×7
- the dihedral angle 90.000000° is one rational part of a half turn plus a whole multiple of the tetrahedron's angle ×5
- a side of 52° is longer than twice the leg, so the half-turn point exists ×2
- the number of sides is a whole number between 3 and 8 ×2
- A is on the circle at the legs' distance beyond the midline
- A is on the sphere
- a kite: and vertically
- a kite: the outward normals weighted by edge length cancel horizontally
- a larger quadrilateral has a larger fourth angle
- a parallelogram: and vertically
- a parallelogram: it has a centre of symmetry, so its edges pair off and the quantity vanishes
- a parallelogram: the outward normals weighted by edge length cancel horizontally
- a polygon of 5 sides falls into 3 triangles
- a polygon of n sides falls into n − 2 triangles
- a rectangle: and vertically
- a rectangle: it has a centre of symmetry, so its edges pair off and the quantity vanishes
- a rectangle: the outward normals weighted by edge length cancel horizontally
- a regular hexagon: and vertically
- a regular hexagon: it has a centre of symmetry, so its edges pair off and the quantity vanishes
- a regular hexagon: the outward normals weighted by edge length cancel horizontally
- a trapezium: and vertically
- a trapezium: the outward normals weighted by edge length cancel horizontally
- a triangle: and vertically
- a triangle: the outward normals weighted by edge length cancel horizontally
- a triangle's quantity does not vanish, so no number of cuts slides it into a rectangle
- A′ is on the circle at the legs' distance beyond the midline
- A′ is on the sphere
- A′′ is on the circle at the legs' distance beyond the midline
- A′′ is on the sphere
- A′′BC has the area of ABC
- A′BC has the area of ABC
- ABC has the area of ABC
- all six pieces have the same edge lengths
- an L: and vertically
- an L: the outward normals weighted by edge length cancel horizontally
- and at Q
- and equal bases, so they are congruent
- and exactly one step up
- and so is its area from its corners
- and so is the angle at P
- and the arc from C
- and the ledger balances: 4 × 6 − 12 = 12
- and the other side at its midpoint
- and the same two tile the square
- and through the point opposite C
- and together they are the triangle's whole angle sum
- and volumes summing to 2
- at half a turn the three pieces tile the rectangle
- B and C are the same distance from the midline
- between one and five swing angles, each between 0 and 180 degrees
- between one and three side lengths, each under 150 degrees
- between two and five shapes
- between two and six sizes for the table, each up to 60 degrees
- each corner is a latitude and a longitude, both within 80 degrees
- each corner tetrahedron has edge 1
- each numerator over its power of three is the cosine of that multiple
- each piece has a sixth of the cube's volume
- each shape is a named list of between three and ten finite points
- each small one's is 6 ⊗ θ
- every angle of each piece is a rational part of a half turn, so its Dehn invariant is nought
- every drawn point is well on the side of the sphere facing the reader
- every edge of a closed solid belongs to exactly two faces
- every labelled point is on the side of the sphere facing the reader
- M is the midpoint of AB
- N is the midpoint of AC
- no edge has zero length
- no multiple of the tetrahedron's angle is a whole number of half turns
- no numerator in the sequence is a multiple of three
- P is the midpoint of A′′B
- P is the midpoint of A′B
- P is the midpoint of AB
- so does the arc from B
- the amount by which the fourth angle exceeds a right angle is the quadrilateral's area
- the angle at O is a right angle
- the apex sits over the base, away from its two ends
- the apex sits over the base, between its two ends
- the arc from A meets the midline at a right angle
- the area of A′′BC from its angles agrees with its area from its corners
- the area of A′BC from its angles agrees with its area from its corners
- the area of ABC from its angles agrees with its area from its corners
- the circle passes through the point opposite B
- the common quadrilateral's summit angles are half the angle sum
- the cube's dihedral angle is a quarter turn exactly
- the Dehn invariant of the cube, edge 1, added up edge by edge, is 0
- the Dehn invariant of the one sixth of a cube, added up edge by edge, is 0
- the Dehn invariant of the regular octahedron, edge √2, added up edge by edge, is −12√2 ⊗ θ
- the Dehn invariant of the regular tetrahedron, edge √2, added up edge by edge, is 6√2 ⊗ θ
- the Dehn invariant of the triangular prism, added up edge by edge, is 0
- the directions are sampled between 90 and 2000 times
- the drawn side is between 2 and 60 degrees
- the fan of triangles tiles the polygon
- the four corners and the octahedron fill the large tetrahedron's volume
- the fourth angle is obtuse, so the four corners cannot all be right angles
- the half-turn about M carries B to A
- the large tetrahedron has edge 2
- the large tetrahedron's Dehn invariant is 12 ⊗ θ
- the midpoint of A′′C is on the midline too
- the midpoint of A′C is on the midline too
- the midpoint of AC is on the midline too
- the number of multiples tested is a whole number between 6 and 30
- the number of risers in the staircase is a whole number between 1 and 5
- the octahedron has edge 1
- the octahedron has the volume of four small tetrahedra
- the octahedron's is −12 ⊗ θ
- the pieces are pulled apart by between nought and one and a half
- the pieces are pulled apart by between nought and one cube
- the polygon is turned by at most a half turn
- the polygon's radius is between 0.5 and 1.5
- the polygon's radius is between a half and two
- the quadrilateral's area from its angles is the triangle's
- the quadrilateral's two legs are equal
- the quadrilateral's two summit angles are equal
- the rectangle and the square have the same area
- the rectangle has the triangle's area
- the same three pieces tile the rectangle
- the side A′′B has the length asked for
- the side A′B has the length asked for
- the side of the square is a whole number between 2 and 24
- the slice meets one side at its midpoint
- the slide is exactly one step across
- the solid has Euler characteristic two, so its faces close up
- the square's side is divisible by n(n+1) so every corner lands on a whole number
- the three pieces tile the triangle
- the triangle fits well inside one hemisphere, where midpoints and perpendiculars are unique
- the triangle is between 0.4 and 1.6 times as tall as its base is long
- the triangle's area from its angles agrees with its area from its corners
- the triangles' areas add up to the polygon's
- the two pieces tile the rectangle
- the view is one of steps, polygon, dehn, hinge, translate, count, sydler, orthoscheme, tetocta, saccheri, lambert, lexell
- the volume of the cube, edge 1 measured from its corners is 1
- the volume of the one sixth of a cube measured from its corners is 1/6
- the volume of the regular octahedron, edge √2 measured from its corners is 4/3
- the volume of the regular tetrahedron, edge √2 measured from its corners is 1/3
- the volume of the triangular prism measured from its corners is 1/2
- three pieces per triangle
- triangles ADM and BEM have equal legs
- triangles ADN and CFN have equal legs
- two tetrahedra and an octahedron of the same edge have Dehn invariants summing to nought
finite-field
55 kinds of claim · 26 placements
- y² = x³ + 0x + 1 over GF(43) has a count inside p + 1 ± 2√p ×1806
- a curve over GF(101) has exactly 82 points ×41
- in GF(7) the sum of the 0th powers is 0 ×33
- every count the bound allows over GF(5) occurs ×16
- every cubic over GF(5) is inside the bound ×16
- the 9 zeros of a sum of 3 squares over GF(3) are divisible by p to at least the promised power ×12
- over GF(7) no cubic misses p by more than 2√p ×11
- there is a field with 4 elements only if 4 is a prime power ×7
- in GF(7) the sum of the 6th powers is −1 ×5
- the prime is between 7 and 61 ×3
- the size of the field is a whole number between 2 and 9 ×3
- 2 zeros and 2 ones contain no 3 with a sum divisible by 3 ×2
- every 5 residues mod 3 contain 3 with a sum divisible by 3 ×2
- the largest prime is a whole number between 13 and 97 ×2
- 1 − f² is 1 at a zero of f and 0 elsewhere
- a conic y² = x² + b misses p by exactly one
- a ring of composite size has a pair of non-zero elements multiplying to zero
- addition and multiplication are commutative
- an irreducible polynomial of the right degree was found
- and it is exactly the composite ones that leave an element without a reciprocal
- and no curve y² = x⁵ + ax + b by more than 4√p
- and so is the mean fourth power, an eighth
- and that variable's power sums to 0 over GF(3)
- and the fewest is the smallest
- between 3 and 8 variables
- each walk ends within 2√p of zero
- every monomial has a variable whose power is not a positive multiple of p − 1
- every non-zero element has a reciprocal
- multiplication distributes over addition
- multiplication is associative
- no two non-zero elements multiply to zero
- odd primes up to 11
- one to three prime-power fields from 3 to 16
- so the number of zeros is 0 mod 3
- so, since (0, 0, 0) is one, there is another
- some row is divisible by exactly the higher power Ax and Katz promise, and no more
- the drawn points and the count agree
- the expansion is drawn over GF(3), where it fits on a page
- the exponent of a product is the sum of the exponents
- the field is closed
- the largest power summed is a whole number between 8 and 20
- the mean square of the normalised error is the semicircle's, a quarter
- the most points any curve has is the largest whole number the bound allows
- the number being factorised is a whole number between 1 and 1000000
- the number of primitive elements is φ(q−1)
- the powers of a primitive element close back onto one
- the size of the ring drawn beside it is a whole number between 0 and 12
- the theorem is checked for n = 3 and n = 5
- the three-variable form is drawn over GF(3), GF(5) or GF(7)
- the tuples are all counted
- the two-variable form is drawn over a prime from 3 to 11
- the view is one the family draws
- the walk ends at the count less p + 1
- the zeros of x² + y² + z² over GF(5) number a multiple of 5
- x² + y² over GF(7) has only the zero at the origin, since −1 is not a square
hamming-code
54 kinds of claim · 27 placements
- class 000 has as many words as the code has ×8
- class 000 has exactly one lightest word ×8
- a row called exact really divides to a power of two
- above capacity it rises
- above it they take over
- and always reaches what Gilbert and Varshamov guarantee
- and it weighs no more than one, which is what perfection means
- and none of them is zero
- and the climb is not monotone — some length is worse than the one before it
- at distance three the bound is only ever exact one short of a power of two
- below capacity the error falls with length
- below the limit failures vanish with length
- each span is a distance between three and five and a longest length up to eight
- every codeword passes all three parity checks
- every pair of erasures is recoverable, since the distance is 3
- every rate is between zero and one
- every two words of the code found are at least the distance apart
- every word in the space is repaired to a codeword by its own syndrome
- four data bits give sixteen codewords
- more repeats, fewer errors
- most of the probability lies within two spreads of np
- nor Plotkin's, where it applies
- nor the Singleton bound
- one rate below 1 − ε and one above
- some length meets the sphere-packing bound exactly
- some triples are and some are not
- the 128 words fall into eight classes
- the block length is a whole number between 20 and 400
- the chance of failure is below 2 to the −m
- the exact answer never beats the sphere-packing bound
- the flip probability is between 0 and a half
- the Hamming code's rate is above capacity at p = 0.1
- the highlighted codeword is a whole number between -1 and 15
- the lightest non-zero codeword weighs what the closest pair are apart
- the longest word length in the table is a whole number between 5 and 23
- the longest word length searched is a whole number between 5 and 8
- the mean of the distribution is the sum
- the minimum distance is a whole number between 3 and 5
- the minimum distance over all 120 pairs is three
- the minimum distance the bound is taken at is a whole number between 3 and 7
- the number of cosets shown across is a whole number between 4 and 8
- the number of message bits is a whole number between 4 and 30
- the rate at the longest length beats the rate at the shortest, at every distance drawn
- the received word is seven bits
- the repetition code and the Hamming code both meet the bound exactly
- the repetition code meets the bound at n = d
- the search finished rather than running out of budget
- the seven columns of the parity-check matrix are distinct
- the seven single errors give seven different syndromes
- the sum settles on the constant
- the syndrome of a single error is the column of the position it hit
- the trials agree with the exact product
- the view is one of syndrome, cosets, bound, best, rate, capacity, typical, repetition, random, rank, erasures, becsim, overhead
- the word the syndrome repairs really is a codeword
incidence
94 kinds of claim · 41 placements
- blocks 0 and 1 meet in exactly one point ×210
- points 0 and 1 lie on exactly one block ×210
- points 0 and 1 share 1 blocks ×156
- the pair 0,1 appears once ×105
- GF(9): 1 has exactly one inverse ×31
- point 0 is in 4 blocks ×20
- the difference 1 occurs exactly once ×20
- the diagonal of the Gram matrix is the block count at point 0 ×13
- PG(2, 5) has n² + n + 1 lines ×7
- PG(2, 5) has n² + n + 1 points ×7
- PG(2, 5): every two points lie on a line ×7
- PG(2, 5): two points lie on only one line ×7
- the parabola y = x² closes to an arc in order 3 ×6
- in odd order 3 no other monomial graph is an oval ×4
- in PG(2, 3) the ovals number q⁵ − q², the count of non-degenerate conics ×2
- no set of q + 2 points of PG(2, 3) avoids three on a line ×2
- PG(2, 2) has arcs of q + 2 points ×2
- a construction is known here for this many points
- a design on this many points was actually found by search
- a failing configuration was found
- a projective plane over GF(q) has q² + q + 1 points
- and distributes from the right
- and each is a conic with its nucleus
- and fails in the nearfield plane
- and no pair differs by nothing
- and none larger
- and therefore there are at least as many blocks as points
- any two lines meet in exactly one point
- but not from the left, and it is not commutative
- Desargues never fails in the plane over GF(9)
- each point is in more blocks than it shares with any other
- each point lies in (v−1)/2 triples
- enough configurations were drawn
- every block's size times the block count is the point count times r
- every line carries q + 1 points
- every member of the base block is a residue modulo n
- every other point off the oval lies on exactly one
- every oval of PG(2, 5) lies on a conic
- every pair of points appears
- every point lies on q + 1 lines
- every tangent passes through the horizontal direction
- exactly one line — the line at infinity — passes every test
- exactly one line passes through any two points
- in every other position it fails on a large share
- in order eight the exponents 2, 4 and 6 give hyperovals
- no point off the oval lies on exactly one tangent
- no three of the points are collinear
- one point lies on every tangent: the nucleus
- q(q+1)/2 points lie on two tangents
- q(q−1)/2 points lie on none
- so the Gram matrix is invertible and the incidence matrix has full column rank
- the base block is between three and six distinct whole numbers
- the design drawn is one that exists
- the design has v(v−1)/6 triples
- the design is one the family knows
- the determinant computed by elimination is the closed form
- the divisibility test and the residue rule agree
- the drawing is the projective plane over GF(2), up to relabelling
- the field order is one of 2, 3, 4, 5, 7, 8, 9
- the field plane never fails (axis at infinity, centre off it)
- the field plane never fails (axis at infinity, centre on it)
- the field plane never fails (finite axis, centre off it)
- the field plane never fails (finite axis, centre on it)
- the forced corners are new points off the axis
- the largest number of points in the table is a whole number between 9 and 25
- the largest number of points tested is a whole number between 9 and 40
- the line at infinity is the last line built
- the nearfield plane has n² + n + 1 lines
- the nearfield plane has n² + n + 1 points
- the nearfield plane: every two points lie on a line
- the nearfield plane: two points lie on only one line
- the nearfield's multiplication is associative
- the number of points in the design drawn is a whole number between 7 and 15
- the number of points in the design is a whole number between 3 and 21
- the order is 3, 4, 5 or 7
- the order is 3, 4, 5, 7 or 8
- the plane is built over a prime between 2 and 5
- the plane is built over a prime field between 2 and 5
- the plane over GF(9) has n² + n + 1 lines
- the plane over GF(9) has n² + n + 1 points
- the plane over GF(9): every two points lie on a line
- the plane over GF(9): two points lie on only one line
- the points are labelled or not
- the residue rule and the divisibility rule agree at every size
- the search agrees with the arithmetic
- the shifts give n distinct blocks
- the size of the permuted set is a whole number between 1 and 8
- the third meeting point is off the line through the other two
- the three meets of corresponding sides are collinear
- the three midpoints are the same distance from the centre, so one circle holds them
- the two divisibility conditions are exactly v ≡ 1 or 3 mod 6
- the view is one of matrix, steiner, sizes, fisher, difference, desargues, nearfield, planecheck, desfail, affineconic, ovals, tangents, monomials, orders, axes, translation
- there are exactly q + 1 tangent lines, one at each point of the oval
- with the axis at infinity and the centre on it, the nearfield plane never fails
latin
172 kinds of claim · 35 placements
- the squares for 1 and 2 produce every ordered pair once ×15
- the count at order 1 is a positive number ×8
- a symbol was found for column 1 ×6
- and once in each column of the square for 1 ×6
- column 1 repeats no symbol ×6
- column 1 was given a symbol it could take ×6
- every symbol appears once in each row of the square for 1 ×6
- relabelling left the pair 1, 2 orthogonal ×6
- symbol 0 is missing from exactly n−k of the columns ×6
- a field of order 4 has 3 non-zero multipliers ×5
- the cyclic square of order 3 has 3 transversals ×5
- the relabelled square for 1 starts 0…3 along its first row ×5
- there is a field with 4 elements only if 4 is a prime power ×5
- row 1 uses every symbol once ×4
- every point lies on 4 lines, one from each class ×3
- the number of rows already placed is a whole number between 1 and 4 ×3
- the 3 available symbols are all used, so an extra square would repeat one ×2
- the order enumerated in full is a whole number between 1 and 4 ×2
- the order of the squares is a whole number between 3 and 8 ×2
- a square laid over itself produces only its own diagonal of pairs
- A₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- A₄: every row of the table is a permutation
- A₄: the operation is associative
- addition and multiplication are commutative
- an associative Latin square has an identity
- an even order from 4 to 8
- an irreducible polynomial of the right degree was found
- an odd order from 3 to 9
- and 9,408 of order 6
- and every table whose elements are all self-inverse has one
- and fifty-six of order five
- and four have every element its own inverse
- and its largest partial transversal has 5 cells
- and nine thousand four hundred and eight of order six
- and no pair of points is met by two lines
- and none of them carries the symbol already standing in that column
- and once in each column of the first square
- and once in each column of the second square
- and once in each column of the square for α
- and once in each column of the square for α+1
- and sixteen of them associate
- and three of them associate
- and twelve have an element of order four
- at order 11 the formula is within five per cent
- D₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- D₄: every row of the table is a permutation
- D₄: the operation is associative
- D₅: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- D₅: every row of the table is a permutation
- D₅: the operation is associative
- D₆: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- D₆: every row of the table is a permutation
- D₆: the operation is associative
- Dic₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- Dic₃: every row of the table is a permutation
- Dic₃: the operation is associative
- each class holds n lines
- each column may still take n−k symbols
- each square either associates or does not
- every line holds n of the points
- every non-zero element has a reciprocal
- every ordered pair of symbols occurs exactly once
- every pair of points is met by some line
- every set of columns can still take at least as many symbols as it has columns
- every square of order 5 has a transversal
- every square of order 6 has a partial transversal of 5 cells
- every symbol appears once in each row of the first square
- every symbol appears once in each row of the second square
- every symbol appears once in each row of the square for α
- every symbol appears once in each row of the square for α+1
- four of the sixteen have every element its own inverse
- multiplication distributes over addition
- multiplication is associative
- no table with an element of order four has a transversal
- no two non-zero elements multiply to zero
- no two pieces share a wrapped diagonal
- no two standardised squares carry the same entry under the corner
- one shift per row already placed
- Q₈: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- Q₈: every row of the table is a permutation
- Q₈: the operation is associative
- relabelling left the pair 1, α orthogonal
- relabelling left the pair 1, α+1 orthogonal
- relabelling left the pair α, α+1 orthogonal
- S₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- S₃: every row of the table is a permutation
- S₃: the operation is associative
- sixteen tables of order four associate
- some square of order 6 has none
- the cyclic square has transversals exactly at odd orders
- the cyclic square of even order 6 has no transversal
- the drawn order agrees with the table
- the field is closed
- the four self-inverse tables have the same number of transversals
- the largest order enumerated is a whole number between 3 and 6
- the Latin square view is one of transversal, mols, bound, net, count, extend, quasi, mates, partial, ryser, growth, groups, queens
- the multiplier of the second square is a whole number between 1 and 7
- the neighbouring orders are shown or not
- the new row uses every symbol once
- the number being factorised is a whole number between 1 and 1000000
- the number of columns is a whole number between 3 and 6
- the number of parallel classes drawn side by side is a whole number between 2 and 5
- the number of squares drawn side by side is a whole number between 1 and 4
- the order of the field the squares are built from is a whole number between 3 and 8
- the order of the plane is a whole number between 2 and 4
- the order of the square is a whole number between 3 and 7
- the pair is built from a prime order, where the construction works
- the parallel classes are the rows, the columns and one per square
- the placements are exactly the transversals
- the quasigroup view is one of counts, tables
- the quoted orders are orders between 7 and 11 and larger than the ones enumerated
- the radius of the points is a whole number between 3 and 8
- the relabelled square for α starts 0…3 along its first row
- the relabelled square for α+1 starts 0…3 along its first row
- the second square uses a multiplier that makes it a different square
- the size of the field is a whole number between 2 and 16
- the size of the permuted set is a whole number between 1 and 8
- the squares for 1 and α produce every ordered pair once
- the squares for 1 and α+1 produce every ordered pair once
- the squares for α and α+1 produce every ordered pair once
- the total at order four agrees with every square of order four, listed
- there are 56 reduced Latin squares of order 5
- there are 576 Latin squares of order four
- there are four reduced squares of order four
- there are twelve Latin squares of order three
- twelve of them have an element of order four
- two associative squares were found to draw
- ℤ₁₀: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₁₀: every row of the table is a permutation
- ℤ₁₀: the operation is associative
- ℤ₁₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₁₂: every row of the table is a permutation
- ℤ₁₂: the operation is associative
- ℤ₂ × ℤ₂ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₂ × ℤ₂ × ℤ₂: every row of the table is a permutation
- ℤ₂ × ℤ₂ × ℤ₂: the operation is associative
- ℤ₂ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₂ × ℤ₂: every row of the table is a permutation
- ℤ₂ × ℤ₂: the operation is associative
- ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₂: every row of the table is a permutation
- ℤ₂: the operation is associative
- ℤ₃ × ℤ₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₃ × ℤ₃: every row of the table is a permutation
- ℤ₃ × ℤ₃: the operation is associative
- ℤ₃: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₃: every row of the table is a permutation
- ℤ₃: the operation is associative
- ℤ₄ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₄ × ℤ₂: every row of the table is a permutation
- ℤ₄ × ℤ₂: the operation is associative
- ℤ₄: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₄: every row of the table is a permutation
- ℤ₄: the operation is associative
- ℤ₅: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₅: every row of the table is a permutation
- ℤ₅: the operation is associative
- ℤ₆ × ℤ₂: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₆ × ℤ₂: every row of the table is a permutation
- ℤ₆ × ℤ₂: the operation is associative
- ℤ₆: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₆: every row of the table is a permutation
- ℤ₆: the operation is associative
- ℤ₇: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₇: every row of the table is a permutation
- ℤ₇: the operation is associative
- ℤ₈: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₈: every row of the table is a permutation
- ℤ₈: the operation is associative
- ℤ₉: a transversal exists exactly when the Sylow 2-subgroup is trivial or not cyclic
- ℤ₉: every row of the table is a permutation
- ℤ₉: the operation is associative
lcg
85 kinds of claim · 47 placements
- the sampled yield at bias 0.2, depth 1, matches the recurrence ×12
- the XOR of 1 flips is 1 with probability ½(1 − (1 − 2p)^k) ×12
- and output 4, which was not ×11
- all 4 windows of 2 bits occur somewhere in the period ×7
- the recovered rule matches output 0, which was given ×4
- the modulus is a whole number between 16 and 65536 ×3
- the register of 4 bits visits every nonzero state before repeating ×3
- a coin that can land either way
- a multiplier consistent with the outputs exists
- a multiplier is a whole number between 2 and 1048576
- a short vector the modulus annihilates exists
- an even number of flips drawn, so they pair up
- and arrives at p(1 − p) bits per flip
- and it is a de Bruijn sequence: every window of that width, exactly once
- and its neighbouring output bits are correlated
- and so is the second
- and the all-zero window never appears
- and they occur equally often, but for the one the all-zero state would have supplied
- and visits none of them twice
- applying the filter twice returns the word
- each extra level adds, and none passes the entropy
- each of them exactly once
- each window is between one bit and the register's width
- every generator's points fall on a genuine family of parallel lines
- every point of the sequence satisfies the relation exactly
- how many high bits are predicted is a whole number between 1 and 8
- no filter passes the ceiling
- no more rows than state bits
- no short whole-number relation holds between consecutive outputs
- raw, beyond one bit, only three outputs in a row are evenly spread — the three words of state
- raw, the patterns that occur number 2^rank
- tempered the rows are independent and raw they are not
- tempered, every pattern occurs equally often but zero, which is one short
- tempered, likewise
- tempered, the rank is full
- tempering never makes the spread worse at any depth
- the all-zero state is the one the register cannot leave
- the all-zero window is the one that is short, by exactly one
- the criterion and the measured period agree
- the dimension is a whole number between 2 and 3
- the example word is a whole number between 0 and 255
- the filter sends the 256 words to 256 different words
- the first prime is a whole number between 3 and 4000
- the first prime is three modulo four, which the construction requires
- the flips drawn is a whole number between 8 and 48
- the highest tap is the register's own width, or it is a shorter register
- the increment is a whole number between 0 and 1048576
- the linear solve, run on this generator, predicts the next output wrongly
- the modulus is small enough that the arithmetic stays exact
- the multiplier is a whole number
- the multiplier is a whole number between 2 and 1048576
- the multipliers really do differ, by more than a factor of two in spacing
- the number of outputs it then predicts is a whole number between 2 and 16
- the number of outputs the solver is given is a whole number between 3 and 8
- the number of points is a whole number between 16 and 4096
- the number of predictions is a whole number between 50 and 4000
- the number of triples is a whole number between 100 and 4000
- the occupied hyperplanes are consecutive, so the family has no gaps
- the output is balanced to within sampling error
- the output of the sticky source is balanced
- the output shows every nonzero window of the register's width
- the outputs drawn is a whole number between 16 and 96
- the patched sequence is as long as there are windows
- the pattern is eight bits, so it fits a 16 × 16 grid
- the predictor gets essentially every output right
- the register does not repeat before its full period
- the register visits every non-zero state before repeating
- the register width is a whole number between 5 and 14
- the relation involves the middle coordinate, so an edge-on view exists
- the relation involves the second coordinate
- the relation's normal lies in the screen, so the planes are seen edge on
- the second prime is a whole number between 3 and 4000
- the solver is given only a handful of outputs
- the taps are positions inside the register
- the triples lie on a small number of planes
- the twisted generator runs through every non-zero state before repeating
- the two multipliers give different numbers of lines
- the vector really does annihilate every point
- the view is one of compare, period, planes, lfsr, spectral, recover, nextbit, bbs, equi, vneumann, rates, sticky, xor, kdist, patterns, rankgrid, temper, raster
- the width of the register is a whole number between 3 and 6
- three to eight multiplier-and-increment pairs
- two multipliers below the modulus
- two taps
- which is enormously better than guessing the leading bits
- while the independent coin's output shows no correlation
poly-code
32 kinds of claim · 14 placements
- the symbol lost at position 1 came back the same ×7
- the number of symbols transmitted is a whole number between 3 and 12 ×4
- the alphabet is a prime between 5 and 17 ×3
- the number of message symbols is a whole number between 2 and 4 ×2
- and inside the Johnson radius the list stays short
- and the count climbs steeply once the radius passes the Johnson bound
- and they differ in that many places
- and this far from the second
- any k of the n symbols recover the message exactly
- at least k symbols survive, or nothing can be recovered and the figure would be a lie
- at the full length every codeword is inside the ball
- every choice of surviving symbols was tried
- every lost position is one of the positions sent
- every non-zero remainder has a reciprocal mod a prime
- more symbols are sent than the message has
- no position is lost twice
- no received word has two codewords within half the minimum distance
- only the zero message gives the zero codeword
- so at least two codewords are within that radius
- the code has two words differing in all but one place
- the code is shorter than the field and longer than the message
- the code meets the Singleton bound exactly
- the field is a prime between 3 and 31
- the lightest non-zero codeword weighs n − k + 1
- the message is k symbols of the alphabet
- the number of received words examined is a whole number between 40 and 2000
- the received word sits this far from the first codeword
- the survivors recover the message
- the view is one of erase, distance, list, pair
- there are at most as many evaluation points as field elements
- which is further than unique decoding reaches
- with one symbol too few, exactly p messages fit the survivors