I will use "p→q" for the indicative conditional "if p, then q". I will use "p⊃q" for the material conditional "(not p) or q". I will say that "indicatives are material" providing that p→q and p⊃q are logically equivalent for all p and q, where a and b are logically equivalent if and only if it is necessary that (a if and only if b). I will say that p entails q provided that it is necessary that p⊃q.
Almost no philosopher thinks indicatives are material. There are very plausible counterexamples. For instance, suppose it is lightly raining in Seattle and Seattle is not having a drought. Let p be "Seattle is having heavy rain" and let q be "Seattle is having a drought". Then p⊃q, since p is false. But it seems quite wrong to say that if Seattle is having heavy rain, then Seattle is having a drought, so p→q doesn't seem to be true.
I am going to offer some arguments that indicatives are material. Say that → is non-hyperintensional provided p→q and p*→q* are logically equivalent whenever p and p* are logically equivalent and q and q* are logically equivalent. Consider the following two theses:
- For any possible world w: (p at w) → (q at w) if and only if (p→q at w).
- For any predicates F and G, from "Every F is a G" (where "x is an F" is more euphonious way of saying that x satisfies F) together with the assumption that c exists, it logically follows that if c is an F, then c is a G.
- If (1) is true and → is non-hyperintensional, then indicatives are material.
- If (2) is true and → is non-hyperintensional, then indicatives are material.
- If (2) is true, then one has to assign the same truth value as the material conditional does to a number of paradoxical-sounding examples of indicative conditional sentences that are relevantly just like the standard alleged counterexamples to the thesis that all indicatives are material.
Argument for (5): Take my heavy rain and drought in Seattle case. Suppose that as it happens, there is no place where there presently is heavy rain. Let Fx say that x is having heavy rain. Let Gx say that x is having drought. Then all Fs are Gs. (If you think, with Aristotle, that "All Fs are Gs" requires there to be an F, then add the premise that on Venus somewhere right now there is a drought but a very, very brief heavy rain is currently occurring. I will leave out such modifications in the future.) Then by (2), we have to say that if F(Seattle), then G(Seattle):
- If Seattle is having heavy rain, then Seattle is having drought.
We can also use (2) to manufacture a true-antecedent, true-consequent case. Suppose that it is raining in both Seattle and the Sahara. Then the following is a standard alleged counterexample of the true-antecedent, true-consequent sort:
- If it's raining in Seattle, then it's raining in the Sahara.
- If earth is a planet on which it is raining in Seattle, then earth is a planet on which it is raining in the Sahara.
Argument for (3): First we need a special case:
- If p and q are non-contingent and → is non-hyperintensional, then p→q is logically equivalent to p⊃q.
- 2+2=4→2+3=5. (necessary, necessary)
- 2+2=5→2+3=6. (impossible, impossible)
- 2+2=5→ (2+2=5 or 1+1=2 or both). (impossible, necessary)
- 2+2=4→2+2=5. (necessary, impossible)
The argument for (3) is now easy. Observe that (p at w) and (q at w) are non-contingent, even if p and q are contingent. So,
- (p at w) → (q at w) is logically equivalent to (p at w) ⊃ (q at w).
- (p⊃q at w) is logically equivalent to (p at w) ⊃ (q at w).
- (p→q at w) is logically equivalent to (p⊃q at w)
Argument for (4): The most intuitive form of the argument is to assume theism, and let Fx say that x is an omniscient being that knows that p, and let Gx say that x knows that q. Then as long as p⊃q, it will be the case that every F is a G (just think about the four possible truth-value combinations). Hence:
- If God is an omniscient being that knows that p, then God knows that q.
If we don't want to suppose there is a God, let's suppose that numbers and sets exist necessarily. Let P be the singleton set whose only member is p. Let Q be the singleton set whose only members is q. Then, let Fx say that x is greater than zero and x equals the number of truths in P. Let Gx say that x is greater than zero and x equals the number of truths in Q. Then, if p⊃q, it is easy to see that all Fs are Gs, so:
- If one is greater than zero and one equals the number of truths in P, then one is greater than zero and one equals the number of truths in Q.