Proof by contradiction (2 ways of writing it) I have a question regarding proof by contradiction (more like the way of writing it).
Let try to prove an easy proposition : For all integer $n$, if $n$ is even, then $n+1$ is odd.

*

*Negate the whole proposition : There exists an integer $n$ such that $n$ is even and $n+1$ is even.
Since $n+1 = 2k$ for some integer $k$,
$n = 2k-1$, a contradiction.


*Let $n$ be an integer. Assume that n is even.
We want to show that $n+1$ is odd.
Suppose to the contrary that $n+1$ is even.
Then $n+1=2k$ which gives $n = 2k-1$, a contradiction.
So the first way (1) negate the whole thing and get a contradiction. Along the way, there are steps setting up quantifier "there exists".
The second way, first set up the direct proof (there is a step doing for all quantifier). Then proceed to suppose a contradiction later.
The question is : Are they both a valid proof ? Are they both called a contradiction ? Or the first one is the contradiction ?(In this case, what is the second approach called ? Or it is NOT a valid proof ?)
 A: Formally proof by contradiction is following:
$$[(\neg P \Rightarrow Q) \land (\neg P \Rightarrow \neg Q)]\Rightarrow P$$
You first case directly use this formula from scratch. In second you use it from middle. Whenever you use it, imho, you can say, that you use proof by contradiction.
A: Let $P(n)$ be that $n$ is even.  So you want to prove the desired result of: $$\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))$$



*

*Negate the whole proposition : There exists an integer $n$ such that $n$ is even and $n+1$ is even. Since $n+1 = 2k$ for some integer $k$, $n = 2k-1$, a contradiction.


Because a contradiction is derived under the assumption that there exists some integer $n$ where $P(n)\wedge P(n+1)$ holds, therefore the desired result is proven.  A few extra steps may make this clearer.
$$\begin{split}\because\quad& \Big(\exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\Big)\to\bot\\&\neg \exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\\&\forall n\in\Bbb Z~.\neg\big(P(n)\wedge P(n+1)\big)\\\hline\therefore\quad &\forall n\in\Bbb N~.\big(P(n)\to\neg P(n+1)\big)\end{split}$$




*Let $n$ be an integer. Assume that n is even. We want to show that $n+1$ is odd. Suppose to the contrary that $n+1$ is even. Then $n+1=2k$ which gives $n = 2k-1$, a contradiction.


Because, for any integer $n$, we derive a contradiction under the assumption that $P(n+1)$ holds while also under the assumption that $P(n)$ holds; therefore the desired result is proven.
$$\begin{split}\because\quad&\forall n\in\Bbb Z~.\Big(P(n)\to\big(P(n+1)\to\bot\big)\Big)\\\hline\therefore\quad&\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))\end{split}$$
A: This is a useful question.  Allow me to streamline the contrast between the two setups:

*

*Assume (to the contrary) that there exists an integer $n$ such
that $n$ is even and $n+1$ is even.

*Let $n$ be an integer, supppose that $n$ is even, and assume (to the contrary) that $n+1$ is even.

When the two assumptions are negated, statements (1) and (2) become equivalent.
