(2nd version to make explicit my implicit assumptions about A, B and C, and the definitions of the non-logical constants "⊂" and "≡".)

Intuitively, the following modal argument seems valid to me and to other people. However, can it be formally proved valid? Or proved not valid?




∀x, ∀y, ∀z, ((x ⊂ y) ∧ (y ⊂ z) → (x ⊂ z))

∀x, ∀y, ((x ⊂ y) ∧ (y ⊂ x) → (x ≡ y))

◇(A ⊂ C)

◻(B ⊂ C)

∴ ◇(A ≡ B)

Or, in ordinary language, where "⊂" means set inclusion and "≡" means identity:

A, B and C exist

A may be some part of C

B is some part of C

Therefore, A and B may be the same part of C

I'm researching expressiveness of modal logic, i.e. to what extent modal logic can express the kind of assertions people make using some ordinary, informal language.

The particular relation identified in the argument here seems to divide the population into (mainly) two groups, and broadly equal in size, those who accept the argument as valid, those who do not. However, those who do not, appear for the moment unable to articulate a conclusive rationale in favour of invalidity.

The only substantial rationale for invalidity offered so far is to exhibit an interpretation of A, B and C which keeps both premises true but makes the assertoric version of the conclusion, i.e. "Therefore, A is the same as B", false.

However, I take this to be inconclusive since it doesn't preclude other interpretations of A, B and C that make true the assertoric conclusion "Therefore, A is the same as B". So, A and B may be the same. QED.

As I see it, there is nothing in the premises that implies that A and B are necessarily different. This in turn implies that they are possibly the same, which is the conclusion of the argument. QED.

Isn't that good enough?

  • Comments are not for extended discussion; this conversation has been moved to chat. – Philip Klöcking Jan 29 '19 at 16:52
  • If you continue to edit the question to rule out answers, it cannot be answered. The form of an explicit quantification usually takes the form "exists A, B, C such that X and Y imply Z". The therefore makes this strange. There clearly do exist such sets. This is no longer vague. It is now just confusing. – user9166 Jan 31 '19 at 8:19

Using Harry Gensler's inference rules one can show as the OP mentions that the argument is invalid. Gensler defines validity in the expected way (page 3):

An argument is valid if it would be contradictory (impossible) to have the premises all true and conclusion false.

The following argument attempts to show validity:

* 1    ◇(A ⊂ C)
  2    ◻(B ⊂ C)
    [∴ ◇(A ≡ B)
* 3    asm: ~◇(A ≡ B)
  4    ∴ ◻~(A ≡ B) [from 3]
  5    W ∴ (A ⊂ C) [from 1]
  6    W ∴ (B ⊂ C) [from 2]
  7    W ∴ ~(A ≡ B) [from 4]

Since I reached no contradiction I construct a refutation of the argument. This shows the invalidity of the argument. I note there is only one possible world and the premises are true (lines 5 and 6) in that world but the conclusion from line 7 is false when A does not equal B which could occur if A is the null set and B is equal to the non-empty set C.

The OP is aware of this argument but is researching the expressiveness of modal logic. However, would one want such an argument to be valid?

Consider the argument as a syllogism:

All A are C.
All B are C.
Therefore, all B are A.

The middle term, that is the one common to the two premises, is C. However, C is undistributed in both premises since it does not immediately follow "all" (see Gensler, page 10). If this argument were valid it would allow the validity of an argument with an undistributed middle which is currently considered a fallacious syllogistic argument.

Also note that the two premises do not show how one can obtain the relation "≡" from "⊂". The premises seem to be incomplete to prove the conclusion.

The following argument with three additional premises would be valid.

* 1    ◇(A ⊂ C)
  2    ◻(B ⊂ C)
  3    ◻(((A ⊂ B) & (B ⊂ A)) → (A ≡ B))
  4    ◻(A ⊂ B)
  5    ◻(B ⊂ A)
    [∴ ◇(A ≡ B)
* 6  | asm: ~◇(A ≡ B)
  7  | ∴ ◻~(A ≡ B) [from 6]
  8  | W ∴ (A ⊂ C) [from 1]
  9  | W ∴ (B ⊂ C) [from 2]
 10  | W ∴ ((A ⊂ B) & (B ⊂ A)) → (A ≡ B) [from 3]
 11  | W ∴ (A ⊂ B) [from 4]
 12  | W ∴ (B ⊂ A) [from 5]
 13  | W ∴ ((A ⊂ B) & (B ⊂ A)) [from 11 and 12]
 14  | W ∴ (A ≡ B) [from 10 and 13]
 15  | W ∴ ~(A ≡ B) [from 7]
 16  ∴ ◇(A ≡ B) [from 6, 14 contradicts 15]

This argument does not use the original two premises (lines 1 and 2) except to bring them into the possible world W. The three new premises (lines 3, 4, and 5) allow a valid argument to be constructed.

"Fallacy of the undistributed middle" Wikipedia https://en.wikipedia.org/wiki/Fallacy_of_the_undistributed_middle

Gensler, H. J. (2002). Introduction to logic. Routledge.

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  • Comment for the "improved" proof: you have assumed the def of set equality (3); you have assumed that the two sets A and B are equal (4 and 5). What is the "benefit" of the following steps that "proves" that A and B are equal ? – Mauro ALLEGRANZA Jan 29 '19 at 9:03
  • @FrankHubeny I think my problem with your first proof is with taking the possible assumption "3 asm: ~◇(A ≡ B)" as proving the arguent not valid. I take this to be the formal expression of producing a counterexample interpretation of A, B and C, whereby in some interpretations, the premises are true and the conclusion is false, which in assertoric argument makes the argument invalid. Yet, it remains obvious to me that the argument is valid, if by "possible" we mean as in the usual sense that it is possible that p if we don't know that not p. – Speakpigeon Jan 29 '19 at 9:13
  • @MauroALLEGRANZA In premise 4, I assume that A is a subset of B. In 5, I assumed that B is a subset of A. I also need premise 3 to define equality in terms of subset, which is how I read "set inclusion". The conclusion of the valid argument allows me to claim I have a proof that A and B are equal. I left premise 1 and 2 in place to show that these two premises did not aid in reaching the result. – Frank Hubeny Jan 29 '19 at 15:37
  • @Speakpigeon Having a formal derivation shows that an argument is valid. That formal derivation uses rules and is somewhat mechanical. If one can't generate the proof showing the argument is valid then one has to produce a refutation which is just an example of true premises and false conclusion. I think I see your point that claiming something is "possible" should grant it something. However, all I think it provides is that the statement may be true. However, if we allowed the argument to be a valid argument then any statement with a diamond in front of it would be a valid argument. – Frank Hubeny Jan 29 '19 at 15:51
  • Frank, unless I am missing something, with your premises 3-5 you can derive not just ◇(A ≡ B) but ◻(A ≡ B), and you do not need 1,2 at all. – Conifold Jan 29 '19 at 23:59

I think your intuition is that it's initially possible that A=B, and adding the premise that A and B belongs to C makes it more probable. Perhaps your intuition could be accounted for in some kind of probability calculus.

But as it stands the argument is invalid. The fact that A and B both belong to C does not "make it possible" that A=B. Perhaps it was initially possible, perhaps not, but your premises do not change that. If it was a possibility before it remains one. If it was impossible before it still is.

Here is an illustration. Take A=29, B=31 and C={the set of prime numbers}. If your argument were valid, one could show by mere modal logic that it is possible that 29=31 whereas everyone knows that it's strictly impossible.

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If A, B, and C are arbitrarily chosen sets, (there exist A, B, and C -- they are 'variables') there is obviously a world where A = B, unless your premises force them to be different, and yours don't.

If A, B, and C are given sets, (for all A, B, and C -- they are 'arbitrary constants') it may be impossible to construct a world where A = B. We must consider all possible relationships between the three things that are permitted by the premises.

A may be given as "A = C - B", for instance (and C may not be empty). Your other premises don't happen to change that. So the conclusion would be false.

You have chosen a symbolism that does not discern these two cases. So you are going to get different answers from different people.

There is a strong trend that unquantified variables are universal, so mathematicians may assume the opposite convention from what you intended. But it is not really a rule, outside of first-order logic.

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