Simplify ¬(s∧(q∨¬s)) to ¬s∨¬q
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¬(s∧(q∨¬s)) to ¬s∨¬q
= ¬((s∧q) v (s∧¬s)) ( From distributive rule: p ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r))
= ¬((s∧q) v F) ( From contradiction rule p ∧ ¬p = F)
= ¬(s∧q) ( From or simplification rule p ∨ F = p)
= ¬s v ¬q ( From demorgan law: ¬(p ∨ q) = ¬p ∧ ¬q)
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