Guess I'm not good at minding my P's and Q's but I am lost.
Logic Question
Sort:
[ push
(P and not P) premise
P separation
not P separation
[ push
not Q premise
P copy
not not P double tild rule <-------------
] pop
if not Q then not P fantasy
if not P then Q contrapositive
Q modus ponens
] pop
if (P and not P) then Q fantasy
Now replace P with the law of the excluded middle (P or not P). How do things work? I can't just use the double-tild rule if I assume (not(P or not P)) can I? Would this imply that the above proof is not necessarily true?
Jan 2, 2011
0
#4
I don't see what you are getting at. If you assume (P and (not P)), clearly you can infer whatever you like. The same is not true of (P or (not P)). Obviously the separation used above would not be valid with an "or".
Wait... I agree that P and not P implies Q, no matter what P and Q are. But what are you then saying about P or not P? Obviously P or not P does not imply anything non-trivial, so unless Q is always true it does not imply Q.
And yes, you can reduce not (P or not P) to not P and not not P, which reduces to P and not P, i.e. false. So not (P or not P) implies Q, and not (P or not P) implies not Q both hold, whatever Q is.
I guess I'm missing your point here.
(For clarity I bolded the predicates, although one may argue whether it increases or decreases the ease to read the text :P)
Jan 2, 2011
0
#6
I believe the inference (not (P or not P)) implies (not P and not not P) requires the law of the excluded middle, so it is not surprising that using this permits a proof of (P or not P).
I am somewhat confused still, unfortunately.
First, I suppose that one may be correct in saying that P and not P imply Q (although as an aside, this may not follow if one assumes "relevant implication" where if-then statements are restricted to reflect more ordinary causality-any opinions about this?). I suppose further that both of you are pointing out to me that this is not the case with assuming (P or not P) and not(P or not P), such as when Elroch says "The same is not true of (P or (not P)). Obviously the separation used above would not be valid with an "or".". If this is all the case, I can understand this.
Perhaps I should rephrase my first question to: Does assuming (P or not P) and not(P or not P) both imply Q?
Jan 2, 2011
0
#8
Since the conjunction [(p v ~p) & ~(p v ~p)], i.e., [(p or not-p) and not-(p or not-p)], is of the form q and not-q, it is a logical contradiction. But for the material conditional--the conditional defined so that "if p, then q" is false when p is true and q is simultaneously false, but is true in all other cases--a contradiction implies anything at all. So, yes, assuming both the law of the excluded middle and its negation allows you to deduce Q--and also to deduce not-Q, and also to deduce that the Moon is made of green cheese, and also to deduce that unicorns can fly, and so on. This is known as ex contradictione quodlibet, or the principle of explosion, and has caused philosophers headaches, leading to attempts to reformulate the conditional in some other way, relevance (as you suggest) often being integral to the reformulation. If you do not intend the material conditional when you say "if p, then q," then you'll have to tell us what form of conditional you do intend.
By the way, your contrapositive step is incorrect. From "if not-q, then not-p," one cannot derive "if not-p, then q." Rather, one derives "if not-not-p, then not-not-q," from which he further derives by double negation cancellation "if p, then q"--not "if not-p, then q." I assume that was merely a typographical error.
Jan 3, 2011
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#9
In answer to strangequark's question at the end, yes. Since it matters not a jot what R is to the fact that (R and not R) implies anything you like, you don't need to worry whether a particular choice of R will do.
This is all clear now. Yes, MindWalk i typed that step wrong. But how can i be so sure that p and not p always imply q? My confusion was that to do this in the case of the law of the excluded middle, that i would come to a point in the proof where i cancel 2 tilds. And we all know that if p then not not p reduces to the law of the excluded middle. So when i assume not(p or not p), how is this valid to do? It does not seem obvious to me that i will arive at q from this....
Jan 19, 2011
0
#11
Here's a proof that (p and not-p) always materially implies q:
- p and not-p (premiss)
- p (1, simplification)
- not-p (1, simplification)
- p or q (2, addition)
- q (4, 3, disjunctive syllogism)
The material implication (if p, then q) is false only if p is true but q is false. It is true the rest of the time--even when p is false. Since (p and not-p) is *always* false, the material implication (if (p and not-p), then q) is always true.
That a material conditional with a false antecedent is always true is known as the principle of explosion (or ex falso quodlibet, if you like Latin phrases). It is a source of some discomfort among philosophers, and a great deal of prose is devoted to finding other ways of construing the English phrase "if p, then q." But the material conditional is what logicians and mathematicians use, and with it, a contradiction implies anything at all.
Jan 21, 2011
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#12
Also, strangequark, note that since, for any A and B,
A and B => A or B (rather trivially),
if you let B be (not R), you get
(R and not R) => (R or not R)
So the law of the excluded middle is very easily proven for any R that satisfies (R and not R).
Jan 24, 2011
0
#13
Well, yes, Elroch, that is so, but it only shows that tautologies are derivable from contradictions. It doesn't show that any statement at all is derivable from a contradiction.
If you use the rules (P -> Q) == (not-P or Q) and (true v P) == (true), then you can easily derive (false -> R) from true, for any R, since (true) == (true v R) == (not-false v R) == (false -> R). Is that what you mean?
Jan 25, 2011
0
#15
MindWalk, strangequark was worried about whether the law of the excluded middle for R was needed in his argument. I pointed out that if you assume the contradition (R and not R) then the law of the excluded middle for R may be derived, so you don't have to worry about whether you are assuming the law of the excluded middle for R. However, strangequark was worried about the law of the excluded middle for (R and not R). i.e. whether (R and not R) or not(R and not R) was provable. Unfortunately, by this stage my main thought is "who cares". Sorry. 
Things that are derived from premises that are not false and not contradictions are inherently more interesting, I feel.
We know that assuming P and not P implies Q. But if we affirm and deny the law of the excluded middle on one hand we seem to get (P or not P) implies not not Q which I assume in this mode we can just say reduces to Q. But using not(P or not ) implies not not Q, which I am not sure reduces to Q without bringing us to the question all over again, so to speak. Thoughts?
Lewis Carroll question: For a modus ponens argument, 1. If A then B 2. A therefore 3. B. one of course has to prove 1 by saying 1. If C then D 2. C therefore 3. D, which leads to an infinite regress...solution?