Finding an interpretation under which Axiom 2 of Aristotelian syllogistic is False and the other axioms True (ignoring Axiom 3, which is derivable from the other axioms). This will require a domain of at least two objects.

In the domain {0,1} there are four ordered pairs: <0,0>, <0,1>, <1,0>, and <1,1>.

Note first that Axiom 6 demands that any member of the domain not in the set assigned to E must be in the set assigned to I. Thus if the set {<0,0>} is assigned to E, then the set {<0,1>, <1,0>, <1,1>} must be assigned to I. Similarly for Axiom 7.

Note secondly that Axiom 3 demands that be a member of the set assigned to E if is. Thus if <1,0> is a member of the set assigned to E, than <0,1> must also be a member. Similarly Axiom 5 demands that be a member of the set assigned to I if is a member of the set assigned to A (but not conversely). Thus if <1,0> is a member of the set assigned to A, than <0,1> must be a member of the set assigned to I.

The problem now is to make Axiom 2 False without falsifying Axiom 1 as well. The solution requires some experimentation.

Here is one interpretation (call it I) under which all the Axioms except Axiom 2 are True:

Domain: {0,1}

A: {<1,0>}

E: {<0,0>, <1,1>}

I: {<0,1>, <1,0>}

O: {<0,0>, <0,1>, <1,1>}

It’s easy to see that Axioms 4 through 7 are True under I. Is Axiom 1 True under I?

Well, Axiom 1 says that if ∈ A and ∈ A, then ∈ A. But under the above interpretation at least one of the conjuncts of this conditional’s antecedent always comes out False. Hence, Axiom 1 is T under I.

Is Axiom 2 False under I?

Axiom 2 say that if ∈ E and ∈ A, then ∈ E. Thus, by Axiom 2 if <0,0> ∈ E and <1,0> ∈ A, then <1,0> ∈ E. But now both conjuncts of the antecedent are True and the consequent is False, so Axiom 2 is False under I.