# In an indirect proof, is it possible to reject the assumption based on contradiction of a premise?

If we have a premise, $x \in A$, is the following a correct use of RAA?

Suppose $x \in A$:

Assume P:

...

Derive $x \notin A$

Therefore not P.

Can we correctly reject P if we find a contraction against the premise when assuming P?

• If $x\in A$ is certain , then we can show that $P$ cannot hold by showing that the assumption that $P$ holds leads to a contradiction, which you did. By the principle of the excluding middle, $P$ must be false. – Peter Sep 16 '18 at 14:12
• The abbreviation RAA is not familiar to me. Perhaps spelling it out (reductio ad absurdum?), or linking to or citing its source would be helpful. – hardmath Sep 16 '18 at 14:51

This is a proof that $x\in A \implies \neg P$. It can be simplified, since everything between "Assume $P$" and "Derive $x\not\in A$" is just a proof that $P\implies x\not\in A$, from which by contrapositive we get $x\in A \implies \neg P$, which was to be demonstrated.

You can't in general conclude from this that $P$ is always false, of course.

Yes. Deriving a contradiction from the assumption of $P$ is a proof for $\neg P$.

Although apparently similar, this is not actually Reduction ad Absurdum.

This is the Rule of Negation Introduction.   Unlike RAA this is an intuitionistically valid rule of inference. $$\dfrac{\Sigma,P\vdash \bot }{\Sigma\vdash \neg P}{\sf\small\neg I}$$

RAA consists of Negation Introduction and Double Negation Elimination. The later rule is not considered valid under an intuitionistic proof system.$$\dfrac{\dfrac{\Sigma, \neg P\vdash \bot}{\Sigma\vdash\neg\neg P}{\small\neg\sf I}}{\Sigma\vdash P}{\small\neg\neg\sf E}$$