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Argument from contradiction

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Original Question
Hi all,

So I think I get this argument.
p or ~p

If I can prove a contradiction stems from ~p then p is true?

Correct? Does it prove p?

Answers

3
Costello tells Abbott he can prove he is not here and Abbott says "Nonsense!"
Costello asks Abbott whether he's in Chicago, New York, San Francisco, and Abbott replies "no".
Costello then says if he's not in Chicago, New York, San Francisco, then he must be somewhere else, right? and Abbott agrees.
Costello then says if he's somewhere else, he's not here.
Disproving ~P does not prove P, without first proving that P+~P=everything

If p and ~p are truly the only options, disproving ~p confirms p.  I suppose we can all imagine fallacious arguments - false dichotomy comes to mind - that attempt to use p=~~p, but as stated in this question I don't see how it can be refuted.


In the real world, my question about a possible third choice, whereby ~p<>~p' would attempt to clarify the apparent dichotomy so that ~~p=p was valid (ie, there was no different ~p')


p: the rocket goes up


~p: the rocket goes down


~p': the rocket did not move

Hi, Wootah!


It seems to me that you are right, Wootah. Here is my argument in defense of your position. 


 


1. If not P, then a contradictory statement is true.
2. No contradictory statements is true.
3. Therefore, not not P. [1, 2 Modus Tollens]
4. If not not P, then P. 
5. Therefore, P. [3, 4 Modus Ponens]


 


Premise 1 is what the original post stipulates. Notice that basic premises 4 in my argument is equivalent to “P or not P”. So, my argument does capture the question asked in your original post. Also, premise 4 is a tautology—it is true in every valuation. And basic premise 2 is grounded in the law of non-contradiction—which is true in every valuation. In other words, the basic premises 2 and 4 are necessarily true. 


Premise 1 is the only basic premise that remains. If it is true, the above argument is sound. That is to say, as long as ‘not P’ really does lead to a contradiction, which is what the OP stipulates, then it will be the case that P is true. If you have any questions about my argument (or think it is not a good argument), please let me know!


Thank you, Wootah.


From, Kaiden

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