Equivalence of contrapositive and contradiction proofs with quantifiers

Question

I have read that contraposition proof is a special case of contradiction proof. For example, the conditional statement: $P \rightarrow Q$, both proofs suppose $\neg Q$. If we show the contradiction $P \wedge \neg P$, then both proofs are equivalent. But I get confused when I introduce quantifiers.

If I would like to proof:

$$(\forall x \in A)(P(x) \rightarrow Q(x))$$

Then, for the contrapositive proof: $$(\forall x \in A)(\neg Q(x) \rightarrow \neg P(x))$$

we suposse $\neg Q(x)$ and try to get $\neg P(x)$

and for the contradiction proof: $$(\exists x \in A)(P(x) \wedge \neg Q(x))$$

The quantifiers are not the same. So, Contraposition is a particular case of contradiction?

Answer 1

Contraposition on the implication inside the statement is a special case of proof by contradiction where one assumes $\neg Q$ and $P$, derives a contradiction (not necessarily to $P$ itself, but some other statement) and concludes $\neg P$. One then has proved $\neg Q \to \neg P$, and by logical equivalence thereby $P \to Q$.
Proof by contradiction on the whole quantified statement means assuming the statement's negation (the existential statement), deriving a contradiction and concluding the positive statement.
Proof by contraposition on the implication is an instance of proof by contradiction, but it is not identical to proof by contradiction on the universal statement. They are different proofs on different formulas and therefore their assumptions look different.

Answer 2

Proof of the contrapositive isn’t really a special case of proof by contradiction. Although in both we prove a statement that is logically equivalent to the desired statement and then appeal to that logical equivalence, the two kinds of argument have different structures. When we prove $p\to q$ by contraposition, we prove the logically equivalent statement $\neg q\to\neg p$. When we prove it by contradiction, we prove the very different statement $p\land\neg q\to\bot$; that is, we assume $p$ and $\neg q$ and derive a contradiction.

They do have in common that in both we assume $\neg q$; the difference is that in one we then prove $\neg p$ directly, while in the other we show that further assuming $p$ leads to a contradiction. The notion that proof by contraposition is a special case of proof by contradiction probably arises from the fact that proofs by contraposition can be presented in the form of proofs by contradiction: one assumes $p$ and $\neg q$, derives $\neg p$ from $\neg q$ without making any use of $p$, and argues that the contradiction $p\land\neg p$ shows that $q$ must have been true after all. This is technically a proof by contradiction, but organizing it as such was completely unnecessary: the guts of it was a direct proof of the contrapositive $\neg q\to\neg p$.

On a final note, $\forall x\in A\,\big(\neg Q(x)\to\neg P(x)\big)$ isn’t really the contrapositive of $\forall x\in A\,\big(P(x)\to Q(x)\big)$: the latter isn’t an implication and doesn’t have a contrapositive. You’ve simply replaced part of the formula by the contrapositive of that part, and since it is logically equivalent, no harm has been done. However, there is no reason to expect the result to look much like the result of negating the quantified expression.