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Statement-I: ~(A⇔∼B) is equivalent to AB

Statement-2:  A(~(A~B)) is a tautology.

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By Expert Faculty of Sri Chaitanya
a
STATEMENT- I is True, STATEMENT-2 is True; S TATEMENT-2 is a correct explanation for STATEMENT- I
b
STATEMENT-I is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT- I
c
STATEMENT-I is True, STATEMENT-2 is False
d
STATEMENT-I is False, STATEMENT-2 is True
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detailed solution

Correct option is B

We have A∨[~(A∧~B)]≡A∨[~A∨B]≡(A∨~A)∨B≡T∨B≡TThus, Statement-2 is true.We have≡~[A⇔~B]=∼[(A→∼B)∧(~B→A)]≡∼[(~A∨~B)∧(B∨A)]≡∼[~(A∧B)∧(A∨B)]≡[(A∧B)∨(~A∧~B)]≡[(A∧B)∨(~A)]∧[(A∧B)∨(~B)]≡[(A∨~A)∨(B∨−A)]∧[(A∨~B)∧(B∨−B)]≡[T∧(~A∨B)]∧[(A∨~B)∧T]≡(A→B)∧(B→A)≡A⇔B


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