--- a coinage by Moore. Or 'coinage', rather. Grice interested in paradoxes associated with it. In which case the alleged paradoxes of 'if' ARE real paradoxes of entailment?
absurdum as not instances of the entailment relation. There have - K Campbell - 1965.
"291). Thus: p entails q iff q follows from or is deducible from p. Subsequent attempts to make this more specific have not met with much success. ...
www.informaworld.com/index/751141477.pdf
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Appendix:
From a recent exchange in a forum:
Subject: Imply vs. Entail
Greetings,
I hope someone can help me with the distinction between implication and entailment. (I know there is another thread in this forum on that topic but I was unable to follow some of the comments so, hopefully, starting my own topic will do the trick.)
If anyone can address the issue in the most general terms (not, say, as it applies to some narrow area of philosophy that I may know nothing about), I would really appreciate it.
Thanks.
moonlight
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Total Posts: 618 #2 Posted Nov 21, 2008 - 3:48 PM:
Hi ThetaPhi,
If a logic is sound and complete, then:
...If a sentence S1 entails another sentence S2, then:
.......It means that the implication of S2 by S1 is logically valid.
In other words:
If S1 |- S2
Then |- S1 => S2
As you can see there isn't that much difference between entailment and implication. Both are talking about the same kind of binary relation, but in different ways.
Cordially,
moonlight.
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MoeBlee
aka I. Kabruob
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Total Posts: 1364 #3 Posted Nov 21, 2008 - 4:09 PM:
moonlight wrote:
If a logic is sound and complete, then:
...If a sentence S1 entails another sentence S2, then:
.......It means that the implication of S2 by S1 is logically valid.
No, that is not soundness and completeness at all.
The soundness theorem is:
If a set of formulas G proves a formula P, then G entails P.
The completeness theorem is:
If a set of formulas G entails a formula P, then G proves P.
moonlight wrote:
In other words:
If S1 |- S2
Then |- S1 => S2
No, that is close to the deduction theorem, not to soundness and completeness.
The deduction theorem is:
For a set of formulas G and formulas Q and P,
If Gu{P} |- Q, then G |- Q->P.
Where G is the empty set, this becomes:
P |- Q, then |- Q->P.
And, in that kind of notation,
The soundness theorem is:
G |- P, then G |= P.
The completeness theorem is:
G |= P, then G |- P.
And what you said at first about entailment and validity (which is correct, but is not soundness or completeness) is put in symbols:
S1 |= S2, then |= S1->S2.
Edited by MoeBlee on Nov 21, 2008 - 4:21 PM
ThetaPhi
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Total Posts: 2 #4 Posted Nov 21, 2008 - 7:28 PM:
Thanks to both of you for taking the time to respond. I appreciate it.
And if humanity has ever developed a less helpful dictionary than the one linked to here I'd love to see it, just for the sake of curiosity.
moonlight
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Location: stuck on earth
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Total Posts: 618 #5 Posted Nov 21, 2008 - 7:42 PM:
MoeBlee wrote:
No, that is not soundness and completeness at all. The soundness theorem is ...
Hold on a second: I was NOT defining soundness. I merely stated the part about the logic needing to be sound and complete so as to be able to use alternatively the terms 'provable' and 'valid'.
MoeBlee wrote:
And what you said at first about entailment and validity (which is correct, but is not soundness or completeness) is put in symbols:
S1 |= S2, then |= S1->S2.
Thank you. I never intended my previous message to include any definition of soundness or completeness. That said your notation is equivalent to mine, since by the soundness and completeness theorems (which is why I mentioned them) we have:
|- S iff |= S
and therefore I chose to use the |- whereas you picked the |= symbol. Let me quote then and highlight from my previous message:
If S1 |- S2
Then |- S1 => S2
So although I wrote on 2 lines instead of one, and although I added 'If' at the front instead of omitting it, and although I used '=>' instead of '->', and although I used '|-' instead of '|=' I believe I wrote exactly the same thing as you wrote.
Well, I hope this clarified matters.
Cordially,
moonlight.
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- Ambrose Bierce -
moonlight
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Total Posts: 618 #6 Posted Nov 21, 2008 - 7:45 PM:
ThetaPhi wrote:
Thanks to both of you for taking the time to respond. I appreciate it.
You're very welcome.
moonlight.
--------------------------------------------------------------------------------
All are lunatics, but he who can analyze his delusion is called a philosopher.
- Ambrose Bierce -
muxol
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Total Posts: 1932 #7 Posted Nov 22, 2008 - 10:10 AM:
MoeBlee wrote:
If Gu{P} |- Q, then G |- Q->P.
Where G is the empty set, this becomes:
P |- Q, then |- Q->P.
I imagine you meant "|- P -> Q" as the consequent, otherwise you are lying.
In modern times entailment is taken to be a semantic notion of "following from" while implication is its syntactic counterpart. E.g. in classical logic one says that some set T of sentences entails another sentence P iff every model of T is a model of P, and T implies P iff there is a deduction (in some system which induces the classical deducibility relation |-) from (some finite subset of) T to P.
MoeBlee
aka I. Kabruob
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Total Posts: 1364 #8 Posted Nov 24, 2008 - 10:03 AM:
moonlight wrote:
I was NOT defining soundness. I merely stated the part about the logic needing to be sound and complete so as to be able to use alternatively the terms 'provable' and 'valid'.
I didn't mean to say that you were definining 'soundness'. You asserted:
(*)
If a logic is sound and complete then
If a sentence S1 entails another sentence S2, then:
It means that the implication of S2 by S1 is logically valid".
The antecedent of (*) is "a logic is sound and complete".
The consequent of (*) is "If a sentence S1 entails another sentence S2, then: It means that the implication of S2 by S1 is logically valid"."
But (*) is true because the consequent itself is true anyway. The antecedent is irrelevent to the truth of the consequent, since the consequent is already true by DEFINITION.
Now you also say "logic needing to be sound and complete so as to be able to use alternatively the terms 'provable' and 'valid'." If by, 'alternatively', you mean 'interchangably', then yes, for FIRST ORDER, that holds. However, that thought is not at all captured by (*), since (*) is true merely by the fact that the consequent is true by definition.
your notation is equivalent to mine, since by the soundness and completeness theorems (which is why I mentioned them) we have:
|- S iff |= S
and therefore I chose to use the |- whereas you picked the |= symbol.
(1) Though we PROVE, for first order, the equivalence, we are not to think that the DEFINITIONS might as well be the same, since to do that would be to overlook the profundity of the PROOF that they are equivalent NOT merely by definition. (And I'm not claiming you claim the definitions ar the same.) (2) What you mentioned in this context was just the deduction theorem, which, yes, is the provablity side of the the consequent of (*), but so it is "in other words" the same as (*) only for FIRST ORDER.
Most importantly, to reiterate, whatever difference or equivalence there is between entailment and provability (let alone between entailment and implication, if, depending on one's definitions, there is a difference or equivalence) is not explained by (*).
hurburble
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Total Posts: 43 #9 Posted Nov 26, 2008 - 11:34 AM:
MoeBlee wrote:
If Gu{P} |- Q, then G |- Q->P.
Where G is the empty set, this becomes:
P |- Q, then |- Q->P.
I suggest you get a beginners book in logic to get this right.
If P entails Q then the implication "P->Q" is a theorem, not the other way around.
muxol
yuletide
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Total Posts: 1932 #10 Posted Nov 26, 2008 - 1:49 PM:
hurburble wrote:
I suggest you get a beginners book in logic to get this right.
If P entails Q then the implication "P->Q" is a theorem, not the other way around.
Why are you repeating me? And why are you incapable of recognizing mistakes that are obviously *typos* rather than misunderstandings?
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I like: p |= q iff p => q is analytic
ReplyDelete(where => is material implication)
RBJ
That's very good. Thanks. We should pursue this in "City of Eternal Truth". That bit above, in the post, I thought you would find inidgestible! I quoted it verbatim from the Philosophical Forum, and did not pay good attention to it. But the problems with 'imply' have more of a Gricen RING than a Gricean sense to it.
ReplyDeleteI.e. Griceians THINK that Grice was involved with 'entailment' when he "coined" implicature, but he was not. He was much more favourable towards definitions like Jones's above.
So, we need to consider 'strict implication'? No. I first did read => to mean strict implication, but of course, it's just the horseshoe, as Jones notes. One good thing about the horseshoe (yes, I know, it's a bother to type) is that it occupies just one slot, unlike ->, or even =>, but that's just typographical.
One could distinguish, "p /= q iff it is analytic that p => q" and
"p /= q iff it is THEOREMATIC that p => q".
There's possibly the idea that theorematic truth yields syntactic /-, while analytic truth yields semantic /=, but I'm speaking vaguely and just to prompt issues!
In any case, apparently Moore (I have to double check that p. 291 of his Philosophical Studies) was reacting against Russell.
This is interesting in it being SO CAMBRIDGE. So, if we think of Carnap as following Russell there (qua Cantabrigensis author) that may be a link (even if just as a reaction) with Grice. Again, I'm expressing badly.
I would think in those participations to colloquium, as in this case, Grice would focus on
"there is a king of France" as being ENTAILED (or strictly implied) by "The king of France is bald"
whereas:
"There is a king of France" is only IMPLICATED (and thus NOT strictly impied or entailed) by "The king of France is NOT bald". I like that.
And note that in WoW:P & CI he does refer to this asymmetry and mentions that indeed the affirmative cases involve 'implication or entailment'. I would need to doublecheck the page in WoW, though.