[(p ∨ q) =⇒ r] =⇒ (~q ∧ r)
~[~(p ∨ q) v r] v (~q ∧ r) 1.Implication law
[(p ∨ q) /\ ~r] v (~q ∧ r) 2.Double Negation and DeMorgan law
I'm stuck at this point.
Look at the truth-table
What are you trying to simplify?
From the way it is posted it seems you are proving the implication:
$[(p \vee q) \Rightarrow r] \Rightarrow (\neg q \wedge r)$
I need to simplify the proposition as much as possible so it would be equivalent to the original expression.
Example:
[(p ⇒ q) ⇒ (p v q)] 0. Original expression
[~(~p v q) v (p v q)] 1. Implication law
[(p /\ ~q) v (p v q)] 2. Doble negation and DeMorgan law
[(p /\ ~q) vp) v q 3. Associative law
pvq 4. absorption law
This is hardly a simplification:
$[(p \vee q) \Rightarrow r] \Rightarrow (\neg q \wedge r)$
$\begin{gathered} \neg [\neg (p \vee q) \vee r] \vee (\neg q \wedge r) \hfill \\
[(p \vee q) \wedge \neg r] \vee (\neg q \wedge r) \hfill \\
[(p \wedge \neg r) \vee (q \wedge \neg r)] \vee (\neg q \wedge r) \hfill \\
\end{gathered}$
You can compare this truth-table with this
to see equivalence.
so the last thing I can apply is distributive law?
If the truth table indicates that they are equivalent, does it mean that the exercise was performed correctly?