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Monday, January 07, 2019


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According to Sep a standard def (not the only one) is

'One sentence presupposes another iff whenever the first sentence is true, the second is true, and whenever the negation of the first sentence is true, the second sentence is true.'

My textbook logic may be rusty, but this entails my 2, no?

Also stopping talking and keeping talking are contraries, not contradictories. True negation contradicts.

Also, you don’t solve the pentad by your proposed revision to 2. Suppose p is true. Then we agree q is true. Suppose p is false. If we agree with (3) that it is necessary that p is true or not-p is true, and if p is false, i.e. not true, it follows that not-p is true. According to your 2*, this entails q (for if p is false, not-p entails q, and not-p, therefore q). But p being true and p being false are the only possible cases. In both cases, q is true, therefore q is necessarily true. But it isn’t.

This seems to reflect a blunder on your part!

Also, as I already pointed out, the example you give (keep talking, stop talking) is not propositional negation.

You are right that the negation of a proposition is its contradictory, not its contrary.

But surely your (2) above is a travesty of the idea of presupposition. We agree, I hope, that if S asserts that John's child is asleep, then S presupposes that John has a child.

Here is an instance of (2):

If *John's child is asleep* presupposes *John has a child,* then (either *John's child is asleep* or *John's child is not asleep*) entails *John has a child.*

But the tautology in the parentheses is not the presupposition of *John's child is asleep.*

The problem may be that you are trying to explain an irreducibly pragmatic notion in terms that are wholly semantic.

P. F. Strawson says this: "... a statement S presupposes a statement S' in the sense that the truth of S' is a precondition of the truth-or-falsity of S . . ." (Intro Log Theory, 175) Thus the truth of *John has a child* is a precondition of *John's child is asleep* having a truth value. Try this:

2A If p presupposes q, then *p has a truth-value* entails *q is true.*

Note that *p v not-p* does not have the same meaning as *p has a truth-value.* Herein, perhaps, lies your mistake.

Tricky stuff!

I have made absolutely no mistake.

>>2A If p presupposes q, then *p has a truth-value* entails *q is true.*

Still doesn't work. Your problem is (3) which fairly explicitly asserts that p is true or not-p is true. Suppose I say (p) 'Socrates has just started arguing', although Socrates has been arguing for hours or days now. So it is not the case that p is true. Then (3) says that not-p is true, i.e. 'it is not the case that Socrates has just started arguing' is true. But if this entails q is true, and this is the only other case, i.e. p or not-p, then q is not contingent.

>>The problem may be that you are trying to explain an irreducibly pragmatic notion in terms that are wholly semantic.

Possibly, but I was fairly clear that the problem revolves around the definition of presupposition. Nor am I clear what 'irreducibly pragmatic' means. The principle of excluded middle is with us, like it or not.

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