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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!

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