Prove that (a ∧ (b → ¬a)) → ¬b is a tautology.
(a∧(b→¬a))→¬b≡(a∧(¬b∨¬a))→¬b≡((a∧¬b)∨(a∧¬a))→¬b≡((a∧¬b)∨F)→¬b≡(a∧¬b)→¬b(a ∧ (b → ¬a)) → ¬b \\\equiv (a ∧ (¬b∨ ¬a)) → ¬b \\\equiv ((a ∧ ¬b)∨(a ∧¬a)) → ¬b \\\equiv ((a ∧ ¬b)∨F) → ¬b \\\equiv (a ∧¬b) → ¬b(a∧(b→¬a))→¬b≡(a∧(¬b∨¬a))→¬b≡((a∧¬b)∨(a∧¬a))→¬b≡((a∧¬b)∨F)→¬b≡(a∧¬b)→¬b
≡¬(a∧¬b)∨¬b≡(¬a∨b)∨¬b≡¬a∨(b∨¬b)≡¬a∨T≡T\\\equiv ¬(a ∧¬b) ∨ ¬b \\\equiv (¬a ∨b) ∨ ¬b \\\equiv ¬a ∨ (b ∨¬b) \\\equiv ¬a ∨ T \\\equiv T≡¬(a∧¬b)∨¬b≡(¬a∨b)∨¬b≡¬a∨(b∨¬b)≡¬a∨T≡T
Thus, it is a tautology.
Hence, proved.
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