I'm reading "Mathematical proofs a transition to advanced mathematics" and in chapter 5 proofs by contradiction are introduced. The introduction though is a little confusing, because it says
"We now introduce a method that can be used to establish the truth of $R$ (which is a proposition) regardless of whether $R$ is expressed in terms of an implication".
but then it explains how to do these kind of proofs like this:
"If $R$ is the quantified statement $\forall x\in S, P(x)\rightarrow Q(x)$, then a proof by contradiction of this statement consists of verifying the implication $¬(\forall x\in S, P(x)\rightarrow Q(x))\rightarrow P∧¬P$"
I don't understand, can this kind of proof technique be used for proving other kind of propositions (theorems)?
An example the book gives is: "Prove that there is no smallest positive real number." and then proceed with a the proof by contradiction. The result to be proved, in predicate logic, would be translated as $¬\exists x \forall y(x<y)$ assuming the domain of $x$ and $y$ is the set of positive real numbers.
So from the example i would say that this proof technique can be used more generally, is this correct?
Remember that implications can be written as disjunctions, $p \implies q \equiv - p \vee q$. So being able to apply it for implications means that you can apply it for disjunctions, which means that you can apply it for conjunctions too (because if you have to prove $p \wedge q$ is true, you can prove $-p \vee -q$ is false).