Supposing
Let’s say I ask you to suppose that p. Naturally, in supposing that p, you won’t at the same time suppose that not-p, even if you’re inclined to believe the latter. But is there anything in the concept of “suppose” such that supposing p means not supposing not-p? I’m inclined to answer `No’. The request that you suppose p and also that not-p might well make sense. Say, I make the request because I want to show you that from a contradiction, anything follows. And if I then request that you suppose that p, without also supposing that not-p, I need not be saying anything redundant. Say, I make this request because the next thing I want to show you is that p makes q very likely, and I no longer need you to suppose that not-p. But the first request won’t make sense, and the second request will be redundant, if the answer is `Yes’. How then do we explain why it’s so natural that when asked to suppose that p, one doesn’t also suppose that not-p?
Assuming ex contradictione quodlibet, it’s not surprising that in the majority of cases, when I ask you to suppose that p, there is a strong implicature that you don’t also suppose that not-p—the point of asking you to suppose that p is defeated if you also suppose that not-p, and that which I want to show follows from p, or is made likely by p, now follows trivially from “p and not-p”.