(1) You can't go wrong in applying hypothetical syllogism (transitivity for conditionals) so long as the set of relevant scenarios for the output conditional is a subset of that for the second input and the set of relevant scenarios for the second input is a subset of that for the first.

(2) You can't go wrong in applying contraposition so long as the set of relevant scenarios for the output conditional is a subset of that for the input.

(3) You can't go wrong in applying strengthening the antecedent so long as the same condition as in (2) is fulfilled.

Let us assume that a conditional A > C is true iff in all relevant scenarios the corresponding material conditional is true.

I will use Sc(X) to denote the set of relevant scenarios for a conditional X .

**Theorem 1.**For any three conditionals A > B, B > C and A > C such that Sc(B > C) ⊆ Sc(A > B) and Sc(A > C) ⊆ Sc(B > C), A > C will be true if A > B and B > C are true.

*Proof:*Take any three conditionals A > B, B > C and A > C such that Sc(B > C) ⊆ Sc(A > B) and Sc(A > C) ⊆ Sc(B > C). Now suppose that A > B and A > C are true. Since all the relevant scenarios for A > C are also relevant for A > B and B > C, and A > B and B > C are true, the material conditionals A ⊃ B and B ⊃ C will both be true at all the relevant scenarios for A > C (by the assumed semantics). Since transitivity holds for material conditionals, A ⊃ C will be true at all the relevant scenarios for A > C, making A > C true (by the assumed semantics). Therefore, if A > B and A > C are true, A > C will be true.

*Illustration*: If I had spoken to a cat then I would have spoken to an animal. If I had spoken to an animal I would have been happy. Therefore, if I had spoken to a cat then I would have been happy. (It is natural to think of the set of relevant scenarios for the first sentence as larger than that for the second. This could be further brought out by adding something like 'no matter what' to the first sentence.)

**Theorem 2.**For any two conditionals A > B and ~B > ~A such that Sc(~B > ~A) ⊆ Sc(A > B), ~B > ~A will be true if A > B is true.

*Proof*: Take any two conditionals A > B and ~B > ~A such that Sc(~B > ~A) ⊆ Sc(A > B). Now suppose that A > B is true. Since all the relevant scenarios for ~B > ~A are also relevant for A > B, and A > B is true, the material conditional A ⊃ B will be true at all the relevant scenarios for ~B > ~A (by the assumed semantics). Since contraposition holds for material conditionals, ~B ⊃ ~A will be true at all the relevant scenarios for ~B > ~A, making ~B > ~A true (by the assumed semantics). Therefore, if A > B is true, ~B > ~A will be true.

*Illustration*: (Assume for the following Q & A that it is analytic that all bachelors are men.)

Q: Do you think that, in view of the fact that we get energy from shooting men, if we don't shoot a man tonight, we won't shoot a bachelor?

A: Of course: If we will shoot a bachelor tonight then, no matter what, we will shoot a man. And what you're asking about follows from that.

**Theorem 3.**For any two conditionals A > B and (A & C) > B such that Sc((A & C) > B) ⊆ Sc(A > B), (A & C) > B will be true if A > B is true.

*Proof*: (Same as for Theorem 2 but with (A & C) > B in place of ~B > ~A, (A & C) ⊃ B in place of ~B ⊃ ~A, and an appeal to the fact that strengthening the antecedent holds for material conditionals in place of the appeal to the fact that contraposition holds for material conditionals.)

*Illustration*: (As with Theorem 2, assume for the following Q & A that it is analytic that all bachelors are men.)

Q: Do you think that we would have shot a man today if we had gone out and shot five bachelors and taken care to go for the masculine-looking ones?

A: If we had gone out and shot five bachelors, then, no matter what, we would have shot a man. So yes, of course we would have shot a man today if we had gone out and shot five bachelors and taken care to go for the masculine-looking ones.

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