Syllogisms that say most

Twenty-four out of two hundred and fifty-six decided the classical syllogism by exhaustion. Each of its sentences — all X are Y, no X are Y, some X are Y, some X are not Y — says something about which regions of a diagram of three overlapping classes are empty or occupied, and a syllogism, two premises and a conclusion about classes $S$, $M$ and $P$, is valid exactly when every diagram satisfying the premises satisfies the conclusion. One thing in each region is enough then showed why exhaustion is finite: if a counterexample exists, one with at most a single thing in each region exists too, so checking the $2^8$ patterns of empty and occupied regions decides everything.

Ordinary argument uses more words than four. Most voters supported the bill; most supporters of the bill were in the governing party; so… — so what? This essay adds most and most … not to the four classical sentences, decides every resulting syllogism, and finds both a new valid form and a familiar-looking argument that fails.

A sentence that needs numbers #

The classical sentences each say that a part of the diagram is empty or that it is not. Most M are P says something different: that the part of $M$ inside $P$ is larger than the part outside. No pattern of empty and occupied regions can express that, because a region marked “occupied” might hold one thing or a million. To decide syllogisms with “most”, the diagram must record how many things are in each region.

So write $n_r$ for the number of things in region $r$, one number for each of the eight regions three classes make. Every sentence becomes a condition on these numbers. All M are P says the regions of $M$ outside $P$ add to nought. Some M are P says the regions of $M$ inside $P$ add to something positive. Most M are P says the regions of $M$ inside $P$ add to more than the regions of $M$ outside it. Each condition is linear: a sum equal to nought, a sum positive, or one sum larger than another.

A syllogism is valid when its premises force its conclusion, which is to say when the premises together with the denial of the conclusion have no solution in non-negative numbers. The denial of a sentence is again linear — the denial of most M are P is that the part inside $P$ is no larger than the part outside — so validity becomes a question of whether a system of linear inequalities has a solution. That is a linear programme, the problem two numbers that have to meet introduced, and it can be decided exactly. Because every condition is unchanged by multiplying all the numbers by the same positive amount, any solution can be scaled to whole numbers, and is a genuine finite counterexample.

Every form, decided #

With six sentence types in each of three places and four figures — the four ways of arranging the middle term $M$ in the premises — there are $6 \times 6 \times 6 \times 4 = 864$ forms. The figures decide each by a small simplex method that maximises the least margin by which the strict inequalities hold; a form is valid when that margin cannot be made positive.

The first check is the classical part. Among the 256 forms using only the four classical sentences, the programme finds exactly fifteen valid — the fifteen moods that are valid without assuming that the classes are non-empty, which is the count the earlier essay found by listing patterns. The method agrees with the old one where the old one applies.

The second is the new part. Sixteen forms with “most” are valid. Some are what one would expect: all M are P, most S are M, so most S are P passes a majority through a universal statement, and no M are P, most S are M, so most S are not P does the same for a negative. The grid shows the overall pattern. Two negative premises yield nothing, as the medieval rules say. Two particular premises — some and some not — yield nothing either. But two premises with most about the same middle term yield a particular conclusion, which no pair of some premises can, and that is the form in the hero figure.

De Morgan’s syllogism #

In the third figure the middle term is the subject of both premises: most M are S and most M are P. The conclusion some S are P follows, by a count. Suppose $M$ has $m$ members, more than $m/2$ of them in $S$ and more than $m/2$ in $P$. If no member of $M$ were in both, the two groups would be disjoint subsets of $M$ with more than $m$ members between them, which is impossible. So at least one member of $M$ is in both $S$ and $P$, and that thing is an $S$ that is a $P$.

Augustus De Morgan published this argument in his Formal Logic of 1847 as an example of what he called the numerically definite syllogism, and used it to argue that the traditional syllogism was a fragment of something larger. He was right, and the larger thing turned out to be quantitative. The hero figure’s counts make the arithmetic visible: ten $M$, six of them $S$, six of them $P$, so at least $6 + 6 - 10 = 2$ in both.

The count generalises at once. If a share $a$ of the $M$ are $S$ and a share $b$ are $P$, at least $a + b - 1$ of the $M$ are both, because the two groups cannot fit side by side inside $M$ once their shares add to more than one. “Most” is the case where both shares exceed a half, which puts the pair in the shaded triangle of the figure; any two shares that add to more than one give the same conclusion, and give it with a size — at least $a + b - 1$ of $M$, not merely some. That is the step from a logic of classes to a logic of proportions that De Morgan and George Boole both tried to take, and it is the step the classical syllogism, with no sizes in it, could not.

Fewer valid forms as ‘most’ is added #

Adding “most” does not add valid forms in proportion; it thins them out.

Fifteen of the 256 classical forms are valid, ten of the 384 with “most” in one sentence, six of the 192 with “most” in two, and none of the 32 with “most” in all three. A syllogism with “most” in both premises and the conclusion is never valid, in any figure. “Most” is a weak connective across a chain, and the most natural argument with it is the one that fails worst.

Most of most is not most #

Most S are M; most M are P; therefore most S are P. It sounds like Barbara, the first and most trusted of the classical moods, with “most” written for “all”. It is invalid.

The counterexample in the figure has four $S$, three of which are $M$, and nine $M$, five of which are $P$ — both premises hold — and not a single $S$ that is $P$. The three $S$ that are $M$ sit in the minority of $M$ that is not $P$. The programme finds such a model automatically as a solution of the premises and the denied conclusion, and scales it to whole numbers; the hand-drawn one is a smaller solution of the same system.

The failure is not marginal. It survives any strengthening of “most” short of “all”.

Ninety-nine per cent of the $S$ are $M$ and ninety-nine per cent of the $M$ are $P$, and still no $S$ is $P$: the ninety-nine $S$ that are $M$ are exactly the ninety-nine $M$ that are not $P$, which is possible because $S$ is a hundred times smaller than $M$. No strength of majority passes along a chain unless the sizes of the classes are also constrained. The same arithmetic governs conditional probabilities. Knowing that $P(M \mid S)$ and $P(P \mid M)$ are both $0.99$ says nothing about $P(P \mid S)$, which can be anything from nought to one, and the essay on evidence that can be multiplied relied on the opposite situation, where conditional statements are combined under assumptions that make them chain.

The opening argument, finished #

The argument the essay began with — most voters supported the bill; most supporters of the bill were in the governing party — is the failed chain. Voters are $S$, supporters are $M$, party members are $P$, and the conclusion that most voters were in the governing party does not follow. Supporters could be a large class of which voters are a small part sitting entirely among the minority of supporters outside the party. If supporters include people who are not voters — lobbyists, say, or members of other bodies — the chain fails in exactly the way the figure draws.

The same arithmetic is behind a paradox in a different subject. The court that contradicts itself found a bench on which a majority accepts each premise of a legal test while a majority rejects the conclusion, because the majorities on different premises are different judges. Majorities do not compose: a majority of a majority is not a majority, and the majorities on two questions need not overlap enough to make a majority on their conjunction — though, by the count in the shares figure, two majorities of one body always share at least one member, so the premise majorities always include some judge who accepts both premises. De Morgan’s syllogism guarantees that judge exists; it does not guarantee that judges like that are a majority, and the paradox lives in the gap between one and more than half. A majority wiser than its members drew the hopeful side of majority arithmetic; this is the sceptical side, and both follow from adding shares.

A diagram that records only occupancy is not enough #

The earlier essays decided the classical syllogism with diagrams that record, for each region, only whether it is empty. That works because every invalid classical form has a counterexample with at most one thing in each region. With “most” the claim fails, and the figures measure by how much.

For every invalid form the figure searches all diagrams with at most $K$ things in each region, for $K = 1, 2, \ldots$, and records the first $K$ at which a counterexample appears. Every invalid classical form fails at $K = 1$, as the earlier essay proved. Of the forms with “most”, 19 need $K = 2$: no diagram with only empty and singly-occupied regions refutes them, but one with a doubled region does. None needs more than two. A typical one is most M are P; some S are M; so some S are P. To refute it, no $S$ may be $P$, so the $M$ that are $P$ must all lie in the region of $M$ and $P$ outside $S$; and since some $S$ is $M$, the region of $M$ and $S$ outside $P$ holds at least one thing. For the $M$ inside $P$ to outnumber those outside, the first region needs at least two things against the second’s one. With single occupancy it can hold only one, and the counterexample disappears.

The 19 are the forms in which “most” must be denied by a tie or a near-tie that single occupancy cannot produce. A diagram that records occupancy cannot see majorities, and so it cannot decide syllogisms with “most”; but a diagram that records counts up to two can, which is a small extension and a real one. The linear programme makes the search unnecessary, since it finds counterexamples of whatever size they need and proves validity by the absence of any.

Erasing, and then counting #

The classical decision procedure in the conclusion is what survives the erasing was graphical: shade the regions the premises declare empty, mark the regions they declare occupied, and read the conclusion off what is left. It works because each classical sentence speaks about regions one at a time — empty or not — and shading and marking are operations on single regions.

“Most” speaks about two groups of regions at once, comparing their totals, and no shading of single regions can represent a comparison. The linear programme is what the erasing method becomes when the marks are replaced by numbers: the premises’ “empty” regions are still erased, their “occupied” regions still required to hold something, and the comparisons are added as inequalities between sums. The classical diagrams, in which four circles cannot do it for four classes and need curves of other shapes, are a geometry for recording which regions are occupied; with “most” the geometry is no longer the point, and the counts in the regions carry everything.

What ‘most’ does to logic #

There is a deeper reason “most” behaves differently, and it goes beyond syllogisms. In the logic of every row or one column — first-order logic, with for all and there exists and relations between things — “most” cannot be defined. Jon Barwise and Robin Cooper proved in 1981 that no first-order sentence expresses most A are B on all finite structures: the comparison of two sizes is beyond the reach of quantifiers that only say “every” and “some”, however many of them are nested. “Most” is a generalised quantifier, one of a family that natural language uses constantly and that first-order logic cannot express.

That makes the decidability of the “most” syllogism a small surprise. Adding “most” to full first-order logic produces a system whose validities cannot be listed by any algorithm. Adding it to the syllogism — three classes, no relations — keeps everything decidable, and the decision is a linear programme. The syllogism’s poverty, its inability to talk about relations, is exactly what makes “most” manageable inside it. Larger fragments of natural-language reasoning with “most”, which allow relations in limited ways, have been studied by Ian Pratt-Hartmann and others, and their decidability depends delicately on what exactly is allowed.

What the computation does and does not establish #

Every one of the 864 forms is decided by an exact criterion — feasibility of a linear system — computed in floating point with a margin far larger than rounding could disturb, and the classical part is checked against the known count of fifteen. The counterexamples are confirmed in whole numbers. The small-model counts are exhaustive searches over every diagram up to the stated size.

What the computation does not establish is anything about readings of “most” other than “more than half”. Ordinary speech uses “most” loosely, sometimes to mean a large majority, sometimes “more than any alternative” among several; each reading is a different linear condition and gives a different table of valid forms, though the method that decides them is the same. And it says nothing about longer chains of reasoning — sorites with “most” in them — where the counterexample of the ninety-nine per cent shows that no fixed threshold survives more than one step.

Still open: how much logic fits in a linear programme #

The syllogism with “most” is decided by linear programming because every sentence it can form is a linear condition on region sizes. Natural language has more: at least three, more S than P, twice as many, exactly half. Each is again linear, and syllogistic fragments built from them are decidable by the same method, but their complexity grows with the number of classes, since the diagram has $2^k$ regions for $k$ classes. Pratt-Hartmann showed that some such fragments with many classes are as hard as linear programming over exponentially many variables, while others are much easier; where exactly the boundary lies between the fragments a reasoner can decide quickly and those it cannot, for the quantifiers people actually use, has been mapped only in part. The single middle term of the classical syllogism keeps everything small, and an argument with “most” and three middle terms is already beyond what any diagram on a page can hold.