We have from truth table for p→qp\to qp→q :
p=1,q=0p=1,q=0p=1,q=0
Then for ¬(p∧q)→q\neg(p ∧ q) →q¬(p∧q)→q :
¬(1∧0)→0 ⟹ 1→0\neg(1\land0)\to 0\implies1\to0¬(1∧0)→0⟹1→0
Answer: false
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