Archangel Macsika
Show ((p ∨ q) ∧ ¬(¬p ∧ (¬q ∨ ¬r))) ∨ (¬p ∧ ¬q) ∨ (¬p ∧ ¬r) is tautology, by using replacement process.
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Solution:
((p ∨ q) ∧ ¬(¬p ∧ (¬q ∨ ¬r))) ∨ (¬p ∧ ¬q) ∨ (¬p ∧ ¬r)
= [(p ∨ q) ∧ ¬(¬p ∧ ¬(q ∧ r))] ∨ [¬(p ∨ q)] ∨ [¬(p ∨ r)]
= [(p ∨ q) ∧ (p ∨ (q ∧ r))] ∨ ¬[(p ∨ q) ∧ (p ∨ r)]
= [(p ∨ q) ∧ (p ∨ (q ∧ r))] ∨ ¬[p ∨ (q ∧ r)]
= (p ∨ q) ∨ T
= T
Hence, it is a Tautology.