Proof By Contra-Positive

Assume $\neg Q$ and conclude $\neg P$. Proof by showing that when assuming the result to be false, there will also be a false condition

Explanation

For a proposition to be fully false, it must have a:

• True condition
• False result If we can prove that the false result actually has a false condition, then our proposition will be saved.

Other explanation

If $A⟹B$, then $\neg B⟹\neg A$

Example

$\forall n\in Z:2^n>n^2⟹n\ge 0$ If, we have a false result - in this case, $n<0$, we must prove that the condition ($2^n>n^2$) is indeed also false.

Proof

$Letn<0$ $⟹2^n=\frac{1}{2^{-n}}<1$ $⟹n^2\ge 1$ This further implies: $⟹2^n<n^2$ This means our condition of $2^n>n^2$ is false. Meaning this result will never be possible with our current condition. QED