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I just had a thought from looking at this answer in this thread: For every $\epsilon >0$ , if $a+\epsilon >b$ , can we conclude $a>b$?

I understand Brian's proof, but I just thought of something. If you prove a hypothesis is untrue, then the logical negation of that hypothesis must be true via proof by contradiction. But why? Can't it be the case that if a hypothesis is false, the negation is also false?

Twenty-six colours
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2 Answers2

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If a hypothesis is false, then the negation of the hypothesis is true. The negation of any false statement is always true, and the negation of any true statement is always false.

It's because a statement can only ever be true or false, there's nothing in between.

The idea behind proof of contradiction is that you basically prove that a hypothesis "cannot be untrue". I.e., you prove that if the hypothesis is false, then $1=0$. You then conclude that it is therefore not true that the hypothesis is false, and in standard logic, that means the hypothesis is true.

More strictly, if you have a hypothesis $H$, and we use the label $\top$ for the trivial true statement and $\bot$ as the trivial false statement, then in a proof of contradiction, you are proving the statement

$$\neg H\implies \bot$$

and you then use the fact that the statement $A\implies B$ is equivalent to $\neg B\implies \neg A$. Using this on your statement, that means that you have proven the statement

$$\neg(\bot)\implies \neg(\neg H)$$

which is the same as

$$\top \implies H.$$

Now, you use the fact that if $A\implies B$ and $A$ are both true, then $B$ is true. So, since $\top\implies H$ is true, and $\top$ is true (by definition), then $H$ must be true.

5xum
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  • Thanks. I might be misunderstanding when there's two or more quantifiers in a statement, so taking the logical negative of the statement confused me – Twenty-six colours May 03 '17 at 07:46
  • @PaulWoch, note that the title of the question that you linked to was indeed formulated incorrectly with respect to quantifiers as noted in a comment there. – Carsten S May 03 '17 at 13:04
  • what is wrong with the following: suppose P is some statement and we prove it by contradiction, then we assume ~P and this leads to some contradiction (e.g. contradicting some axiom). Then, we conclude that things are fine and this contradiction must mean the opposite is true so ~~P=P is true. But what worries me (for some reason doesn’t require proof or justification) is that now, if we start with P, how do we know if we were to follow the same rules of inference and axioms that we won’t arrive again at a contradiction? How is it that this is impossible? What guarantees this doe snot happen? – Charlie Parker Feb 16 '18 at 14:47
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Let's say you want to prove $P \implies Q$.

There are some substeps $P \implies P_2 \implies P_3 \implies P_4 \implies Q$.

Assume that $P$ is true and $Q$ is false.

Since $P$ is true, we have $P_2$ is true.

Since $Q$ is false, $P_4$ is false (by contrapositive of the last implication).

Since $P_2$ is true, $P_3$ is also true.

Since $P_4$ is false, $P_3$ is also false (by contrapositive).

So, $P_3$ is both true and false. This is impossible.

So, when $P$ is true, $Q$ cannot be false.

ThePortakal
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    Since Q is false, P4 is false (by contrapositive of the last implication). and here we use thing we wanna proof/explain

     – RiaD May 03 '17 at 10:37
  • @RiaD "thing we wanna proof/explain" -- is this "thing" that you mention here $P\implies Q$ (the original theorem to be proved) or is it "$P$ is true and $Q$ is false" (the negation of the theorem)? – David K May 03 '17 at 12:44