If S(p,q,r)=∼p ∨∼(q ∨ r) is a compund statement, then S(~p,~q,~r) is
S(p,q,r)=∼p ∨∼(q ∨ r)S( p, q, r)=∼(∼p) ∨∼(∼q ∨∼r)=p ∨∼(∼(q ∧ r))=p ∨(q ∧ r)
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