Comments on Another Round on (Semantic) Presupposition: An Inconsistent PentadTypePad2019-01-07T22:10:00ZBill Vallicellahttps://maverickphilosopher.typepad.com/maverick_philosopher/tag:typepad.com,2003:https://maverickphilosopher.typepad.com/maverick_philosopher/2019/01/another-round-on-semantic-presupposition-an-inconsistent-pentad/comments/atom.xml/The light ostrich commented on 'Another Round on (Semantic) Presupposition: An Inconsistent Pentad'tag:typepad.com,2003:6a010535ce1cf6970c022ad3acbafa200d2019-01-08T13:36:56Z2019-01-08T19:58:33ZThe light ostrichI have made absolutely no mistake. >>2A If p presupposes q, then *p has a truth-value* entails *q is true.*...<p>I have made absolutely no mistake.</p>
<p>>>2A If p presupposes q, then *p has a truth-value* entails *q is true.*</p>
<p>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.</p>
<p>>>The problem may be that you are trying to explain an irreducibly pragmatic notion in terms that are wholly semantic. </p>
<p>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.</p>BV commented on 'Another Round on (Semantic) Presupposition: An Inconsistent Pentad'tag:typepad.com,2003:6a010535ce1cf6970c022ad3869eb9200c2019-01-08T12:27:08Z2019-01-08T19:58:33ZBVYou are right that the negation of a proposition is its contradictory, not its contrary. But surely your (2) above...<p>You are right that the negation of a proposition is its contradictory, not its contrary. </p>
<p>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. </p>
<p>Here is an instance of (2):</p>
<p>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.*</p>
<p>But the tautology in the parentheses is not the presupposition of *John's child is asleep.*</p>
<p>The problem may be that you are trying to explain an irreducibly pragmatic notion in terms that are wholly semantic. </p>
<p>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:</p>
<p>2A If p presupposes q, then *p has a truth-value* entails *q is true.*</p>
<p>Note that *p v not-p* does not have the same meaning as *p has a truth-value.* Herein, perhaps, lies your mistake.</p>
<p>Tricky stuff!</p>The Bad Ostrich commented on 'Another Round on (Semantic) Presupposition: An Inconsistent Pentad'tag:typepad.com,2003:6a010535ce1cf6970c022ad3869837200c2019-01-08T08:09:02Z2019-01-08T11:17:57ZThe Bad Ostrichhttp://logicmuseum.comAlso, you don’t solve the pentad by your proposed revision to 2. Suppose p is true. Then we agree q...<p>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.</p>
<p>This seems to reflect a blunder on your part!</p>
<p>Also, as I already pointed out, the example you give (keep talking, stop talking) is not propositional negation. <br />
</p>Ostrich commented on 'Another Round on (Semantic) Presupposition: An Inconsistent Pentad'tag:typepad.com,2003:6a010535ce1cf6970c022ad3cc365f200b2019-01-07T22:58:27Z2019-01-08T11:17:57ZOstrichAlso stopping talking and keeping talking are contraries, not contradictories. True negation contradicts.<p>Also stopping talking and keeping talking are contraries, not contradictories. True negation contradicts.</p>Ostrich commented on 'Another Round on (Semantic) Presupposition: An Inconsistent Pentad'tag:typepad.com,2003:6a010535ce1cf6970c022ad3cc3622200b2019-01-07T22:52:56Z2019-01-08T11:17:57ZOstrichAccording to Sep a standard def (not the only one) is 'One sentence presupposes another iff whenever the first sentence...<p>According to Sep a standard def (not the only one) is</p>
<p>'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.'</p>
<p>My textbook logic may be rusty, but this entails my 2, no?</p>