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Is the function $(∼p ∨ q) ∧ (p ∨ ∼r) ∧ (∼p ∨ ∼q)$ equal to the function $∼(p ∨ r)$?

Practicing out of a textbook for my test - I honestly have no idea how to simply this algebreically. If I had to do this still I would just pick truth values to assign to p, r, and q in order to find a counterexample to them being logically equivalent ... or construct a truth table (wanna see if there's a less tedious wayy of solving)

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You can use logical equivalences to rewrite the expression:

$(\neg p \lor q) \land (p \lor \neg r) \land (\neg p \lor \neg q) \Leftrightarrow$ (Commutation)

$(\neg p \lor q) \land (\neg p \lor \neg q) \land (p \lor \neg r) \Leftrightarrow$ (Adjacency)

$\neg p \land (p \lor \neg r) \Leftrightarrow$ (Reduction)

$\neg p \land \neg r \Leftrightarrow$ (DeMorgan)

$\neg (p \lor r) $

Bram28
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