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Question: Dual of \((x^{'} \vee y^{'})^{'} = x \land y\) is...

Dual of (x′∨y′)′=x∧y(x^{'} \vee y^{'})^{'} = x \land y is

A

(x′∨y′)=x∨y(x^{'} \vee y^{'}) = x \vee y

B

(x′∧y′)′=x∨y(x^{'} \land y^{'})^{'} = x \vee y

C

(x′∧y′)′=x∧y(x^{'} \land y^{'})^{'} = x \land y

D

None of these

Answer

(x′∧y′)′=x∨y(x^{'} \land y^{'})^{'} = x \vee y

Explanation

Solution

Change ‘∨’ to ‘∧’ and ‘∧’ to ‘∨’.