This is the definition of $ \bf PL $
Let $ S = S _ { \bf PL } $, the set of logical symbols for $ \bf PL $, be the union of the following three sets:
$ Con = \{ \neg , \lor , \land , \rightarrow , \leftrightarrow \} $ is the set of connectives;
$ Props = \{ P _ 1 , P _ 2 , P _ 3 , \dots \} $ is the (countably infinite) set of propositional variables; and
$ \{ ( , ) \} $, the left and right parentheses.
And here are the deduction rules.
I tried to start by assuming $ Q $.
Deduction verifying $ \neg ( Q \rightarrow ( P \lor Q ) ) $
- $ Q ^ * \qquad \qquad \qquad $ Hypothetical
- $ ( P \lor Q ) ^ * \qquad \qquad (VI) $ on $ ( 1 ) $
- $ Q \rightarrow ( P \lor Q ) \qquad ( \rightarrow I ) $ on $ ( 1 , 2 ) $
But I can't find a falsehood. How can I continue?
I doubt I am already doing wrong in step 3. $ Q ^ * $ is already used for constructing $ ( P \lor Q ) ^ * $. If I want to construct $ Q \rightarrow ( P \lor Q ) $, I have to assume $ Q ^ { ** } $ before, isn't it?
Your step $(3)$ is fine. Having assumed $Q$, you derive $Q\lor P$ correctly in step (2). Then, you are justified by concluding (deriving) that $Q\rightarrow (P \lor Q)$ through $\to$ Introduction.
From $(3)\quad Q\to(P \lor Q))$ $\quad (\to I)\quad$ on $(1,2)\;\;$ [at which you've discharged the assumption Q]
you can use $\lnot\lnot\,$I to conclude $$(4)\quad\lnot\lnot (Q\to (P\lor Q))\equiv \lnot (\lnot( Q\to (P\lor Q)))\qquad (\lnot\lnot I)\quad \text{ on } (3)$$
In other words, given your final comment about what you are seeking to do ("Verify ¬(Q →(P ∨ Q)) is a falsehood."),
you will have proven that it is NOT the case that $\lnot(Q\to(P\lor Q))$ is true. In other words, you will have proven that $\lnot(Q\to (P\lor Q))$ is indeed, false.