I’ve long been worried by the following fact. Very often, when (~p or q) is assertable, so is p -> q, and vice versa. This is so widespread that we might worry that it is to be explained by the simple fact that (~p or q) entails p -> q, and vice versa. (I’m using -> for if…then, in case this isn’t obvious.) Now, I don’t want to be making a simple confusion of assertion conditions with truth conditions here. I’m not at all worried by the flat-footed argument that we have a preservation of assertion conditions here, so we must have a preservation of truth conditions. But I do think that when it seems that two sentences have the same assertion conditions, that’s something that needs explaining, and perhaps the best explanation of it is that they have the same truth conditions.

One might respond that p -> q can’t have the same truth conditions as (~p or q) because their negations have different assertion conditions. But an argument that simple would just be a confusion between truth and assertability, and not worth the electrons it’s reflected off.

Why the concern then? Well, largely because I’m in print rejecting the equivalence between p -> q and (~p or q), and I would like the things I’m in print rejecting to be wrong. Not the deepest reason ever, but a reason.

Anyway, here’s a little argument that the equivalence (or near equivalence) in assertion conditions between p -> q and (~p or q) is not grounded in a truth-conditional equivalence. Consider the following argument.

1. (A & B) -> C Premise
2. ~(A & B) or C From 1
3. (~A or ~B) or C DeMorgan from 2
4. (~A or C) or (~B or C) Distribution from 3
5. (A -> C) or (B -> C) From 2

Note that this argument is not assertability preserving. We could be in a position to say, for example, If you fill the room with gas and light a match, the room will explode, while not being in a position to say If you fill the room with gas, the room will explode, or if you light a match, the room will explode. In that case, all the steps before the last one are assertability preserving. Letting A, B and C be as defined in that example, we can say 1, 2, 3 and 4. But the inference to 5 is mistaken.

Now that should be an interesting fact. If (~p or q) is equivalent to p -> q, then we would expect that we could just substitute one for the other in a disjunction. But that’s what we cannot do here. This isn’t a knock-down argument, but it is I think a little evidence that we should be looking for a pragmatic explanation of what (~p or q) and p -> q have in common, rather than what separates them. This is hardly a new conclusion – there’s a Stalnaker paper from I think 1975 trying to explain why the assertion conditions for (~p or q) and p -> q might be the same even though their truth conditions are different – but I don’t think I’ve ever seen anyone argue quite this way for Stalnaker’s approach.