q ∧ ( p → ¬q) → ¬p
q ∧ ( ¬p ∨ ¬q) → ¬p
(q ∧ ¬p)∨ (q ∧ ¬q) → ¬p
(q ∧ ¬p)∨ F → ¬p
i dont know how to solve this further. Kind of leaves me confused what would be the next step
q ∧ ( p → ¬q) → ¬p
q ∧ ( ¬p ∨ ¬q) → ¬p
(q ∧ ¬p)∨ (q ∧ ¬q) → ¬p
(q ∧ ¬p)∨ F → ¬p
i dont know how to solve this further. Kind of leaves me confused what would be the next step
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First, a term like $P \lor F$ is equivalent to just $P$. So, as the next step you get:
$(q \land \neg p) \to \neg p$
And now rewrite this second implication just as you did the first. That is, the next step is:
$\neg (q \land \neg p) \lor \neg p$
Now do DeMorgan and you're almost there!