A B A → B A ∨ B F=(A → B) ↔ (A ∨ B)0 0 1 0 10 1 1 1 11 0 0 1 0 1 1 1 1 1
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A B A → B A ∨ B F=(A → B) ↔ (A ∨ B)
0 0 1 0 1
0 1 1 1 1
1 0 0 1 0
1 1 1 1 1