Proof by contradiction (2 ways of writing it)

Question

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.

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.

2. 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 ?)

Answer 1

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.

Answer 2

Let $P(n)$ be that $n$ is even. So you want to prove the desired result of: $$\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))$$

---

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.

$$\begin{split}\because\quad& \Big(\exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\Big)\to\bot\\&\neg \exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\\&\forall n\in\Bbb Z~.\neg\big(P(n)\wedge P(n+1)\big)\\\hline\therefore\quad &\forall n\in\Bbb N~.\big(P(n)\to\neg P(n+1)\big)\end{split}$$

---

2. 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.

$$\begin{split}\because\quad&\forall n\in\Bbb Z~.\Big(P(n)\to\big(P(n+1)\to\bot\big)\Big)\\\hline\therefore\quad&\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))\end{split}$$

Answer 3

This is a useful question! Here's a blow-by-blow outline.

The two basically equivalent ways to begin proving the statement $$\forall x{\in}S\;\big(P(x)\implies Q(x)\big)$$ using Proof by Contradiction:

• Assume that there exists an element $x$ of set $S$ such that $P(x)$ is true but $Q(x)$ is false. $\quad$ Here, the assumption is $$\exists x{\in}\mathbb S\:\big(P(x)\land \lnot Q(x)\big).\tagα$$
• Take any element $x$ of set $S,$ suppose that $P(x)$ is true; assume that $Q(x)$ is false. $\quad$ Here, the assumption is just $$\lnot Q(x).\tagα$$

Then we want to derive any contradiction, say $$\beta\land\lnot \beta$$ (e.g., by deriving the falsehood $\beta),$ so as to deduce that the foregoing assumption $\alpha$ must in fact be contrary to the truth, i.e., to falsify $\mathbf\alpha,$ thereby obtaining $$\lnot\alpha.$$ Either (following the 1st way) immediately conclude that the original statement is true or (following the 2nd way) use the inference rules Conditional Introduction and Universal Generalisation to arrive at the original statement.

P.S. In general: The structure of a Proof by Contradiction.