Proving a statement of the form $P\rightarrow(\lnot Q\lor \lnot R)$.

Question

Not looking for a proof of my question, only an answer to my question below.

Question: I want to assume that $P$ is true for this proof, then if I assume that $Q$ is also true and conclude that $R$ is not true, does that prove this statement? Do I also have to prove that if $R$ is true then $Q$ is not? Thank you for the help.

For reference the logical form of the statement: Suppose $f:[a,b]\rightarrow\Bbb{R}$ is not constant on $[a,b]$, then $f$ is not continuous or $f([a,b])⊄\Bbb{Q}^c$.

Answer 1

You're right.

This is because $\lnot Q\lor\lnot R$ is equivalent to $Q\to \lnot R$. There is no need to prove $R\to\lnot Q$ if you prove the latter (due to the symmetry of $\lor$).

Also: $$\begin{align} P\to(\lnot Q\lor\lnot R)&\equiv (\lnot P)\lor((\lnot Q)\lor(\lnot R))\\ &\equiv ((\lnot P)\lor(\lnot Q))\lor(\lnot R)\\ &\equiv (\lnot(P\land Q))\lor\lnot R \\ &\equiv (P\land Q)\to \lnot R.\end{align}$$