Does All Imply Some?

So in the logic course I am currently teaching we are about to start talking about the contrast between the modern and traditional square of opposition. The main difference, of course, is that poor Aristotle thought that if it was true that All A’s are B’s that made it the case that Some particular A was in fact B, and so he would be forced to exclude syllogisms as valid that have fictional terms. Consider the following proposition,

1. Some humans are mammals

Now, I often find that students think that 1 is false. They think that it is false because they think that it implies 2

2. Some humans are not mammals

but in the traditional square of opposition this is not the case. 1 and 2 are subcontraries, which means that they can’t both be false. Now this means that it is a possibility that both of them are true, but there is also the possibility that only one of them is true and the other is false.  Since it is true that all humans are mammals by subalternation we know that 1 is true . Thus, we can say that 1 is true because we know that 3 is true and 2 has to be false because it is the contradiction of 3.

3. All humans are mammals

But what are we to make of this on the modern interpretation? Can we no longer say that 1 is true because of the fact that all humans are mammals? It is still the case that 2 is false, as it is still the contradiction of 1. But what about the so-called existential fallacy?

But if we take this discussion out of the realm of categorical propositions and start talking in terms of truth-functional semantics don’t we get the traditional square back? So, 1 is written as 1* with

1* Ex(Hx & Mx)

Whereas 2 and 3 turns into 2* and  3*

2* Ex(Hx & ~Mx)

3* (x)(Hx–>Mx)

So to get the traditional square back we would need to show that 3* implies 1*.

Consider the following proof,

A. (x) (Hx –>Mx) [assumption]

B.  Ex (Hx) [assumption]

C. Hy [Existential Instantiation of B]

D. Hy –> My [Universal Instantiation of A]

E. My (Modus Ponens D, C)

E. Hy & My [conjunction introduction]

F. Ez (Hz & Mz) [existential generalization]

So, if B is correct and there are humans then 3 does entail 1 just like Aristotle said. Am I missing something?

9 thoughts on “Does All Imply Some?

  1. I think that it’s as you say: _if_ anything at all is H, then you can get from (Ax)(Hx -> Fx) to (Ex)Fx. But that extra premise (the (B) in your example) is absolutely crucial. That’s why we don’t say we have the traditional square, because that additional premise is required. (3) by itself doesn’t entail (1) — only (3) and (B) together do.

  2. I thought the whole point was that 3. implied that there were humans in the Aristotelian scheme. So moving over to TFS the whole question would be whether or not A implies B. Aristotle would say yes, moderns say no. Right?

  3. Hi Jonathan and Ben, thanks for the comments!

    Ben, the issue is traditionally put in terms of whether the universal affirmative proposition, when true, implies the singular affirmative proposition (which does imply B).

    Jonathan, I agree that without B the argument doesn’t go through. Without it all you can get is that there is something which is either not a human or is a mammal (EX (~Hx v Mx))…which is just about as good as saying that something or other exists and all we know about it is that it is not the case that it is both Human and non-mammalian….but the idea with B was supposed to be to capture Aristotle’s implicit assumption that the terms are not empty…but I guess you are right…this is really saying no more than that the inference is conditionally valid…

  4. I think writing 3 as 3*, (x)(Hx->Mx), and not as E(x)(Hx->Mx) is already an interpretive step for your part. You read the major premise as an ‘if-then’ conditional and not as a existential statement. Aristotle was wrong in not observing the ambiguity – which permits both translations (of which only the latter implies 1, and here you are right) – but your steps form A to F rest on this stipulating, and it should have been in the foreground.

  5. Hi Argumentics, I am not sure I am following you here. How is (what I will call ‘3#’) supposed to capture the claim that all humans are mammals? That will be true if at least one human is a mammal but it will also be true if only one human is a mammal and the rest aren’t…in fact you can prove that statement quite easily from my 3*, as follows,

    1# (x) (Hx –> Mx)
    2# Ha –> Ma (UI of 1)
    3# Ey (Hy –> My) (EG from 2)

    In logic courses they condition us to not use the conditional in existentially quantified statements so we are taught to write 3# as Ex ~(Hx & ~Mx) but that’s just a stylistic thing…

    Finally, in my proof above 3# can be used just as well as 3* so it wouldn’t matter anyways…

  6. This is just a more technical way of putting Jonathan’s point, but your proof is not valid by the standards of natural deductive proof systems for classical first order logic, since you haven’t discharged your assumption in (2). That’s an important rule, since undischarged assumptions could be used to prove anything from anything!

    There are some ways of doing semantics and proof theory for first order logic that play with the existence assumptions carried by the quantifiers though. Free logics (in which the quantifiers carry no existence assumptions) are the most well known, but there are also some well-explored logics in which the universal quantifier carries existence assumptions (I.e. In which universal instantiation is valid). According to mthere seconds of research, there is a paper by Gentzen and/or Jaskowski from 1934 about this.

  7. Hey Daniel, thanks for the comment and I hope to see you out at the Parkside tonight!

    As a minor note, even if the assumption isn’t discharged the deduction is still valid, it is just that it is only valid conditionally, i.e. on the condition that Ex(Hx)…but to the substance of your point, I don’t see why that isn’t easily avoided thusly: suppose that I introduce premise A0,

    A0. (x) (Hx –> Mx) & Ex (Hx) [assumption]

    Then I would get A and B by conjunction elimination and the proof would proceed as above except that the conclusion would be G

    G. ((x) (Hx –> Mx) & Ex (Hx)) –> (Ey (Hy & My))

    which discharges the assumption. But even so, I think Jonathan’s point still stands. The A proposition doesn’t all by itself imply the I proposition. It only does so when we know that the subject term is not an empty term, which is basically equivalent to rejecting the traditional square (except in cases where we know that the terms are non-empty…)

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