Let us show that ∼((p∨(q→∼r))∧(r→(p∨∼q)))≡(∼p∧q)∧r:
∼((p∨(q→∼r))∧(r→(p∨∼q)))≡∼((p∨(∼q∨∼r))∧(∼r∨(p∨∼q)))≡∼((p∨(∼r∨∼q))∧(∼r∨(p∨∼q)))≡∼((p∨∼r)∨∼q)∧(∼r∨(p∨∼q)))≡∼((∼r∨p)∨∼q)∧(∼r∨(p∨∼q)))≡∼((∼r∨(p∨∼q))∧(∼r∨(p∨∼q)))≡∼(∼r∨(p∨∼q))≡∼(∼r)∧∼(p∨∼q))≡r∧(∼p∧∼(∼q)))≡r∧(∼p∧q))≡(∼p∧q))∧r.
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