Question

A logical binary relation $⨀$, is defined as follows:
$\begin{array}{ccc}A& B& A⨀B\\ True& True& True\\ True& False& True\\ False& True& False\\ False& False& True\end{array}$

Let$\sim$ be the unary negation (NOT) operator, with higher precedence, than $⨀$. Which one of the following is equivalent to $A\wedge B$ ?

• A. $(\sim A⨀B)$
• B. $\sim (A⨀\sim B)$
• C. $\sim (\sim A⨀\sim B)$
  
• D. $\sim (\sim A⨀B)$

Answer 1

The correct option is D $\sim (\sim A⨀B)$

By using min terms we can define.

$A⨀$ B = AB + $AB{}^{\prime }+A{}^{\prime }B{}^{\prime }$
= A + $A{}^{\prime }B{}^{\prime }$
$=(A+A{}^{\prime })\cdot (A+B{}^{\prime })=A+B{}^{\prime }$
(a) $\sim A⨀B=A{}^{\prime }⨀B=A{}^{\prime }+B{}^{\prime }$
(b) $(\sim A⨀\sim B)=(A⨀B{}^{\prime }){}^{\prime }=(A+(B{}^{\prime }\left){}^{\prime }\right){}^{\prime }$
$=(A+B){}^{\prime }=A{}^{\prime }B{}^{\prime }$
(c)$\sim (\sim A⨀\sim B)=(A{}^{\prime }⨀B{)}^{\prime }=(A{}^{\prime }+B{}^{\prime }\left){}^{\prime }\right){}^{\prime }$
= $(A{}^{\prime }+B){}^{\prime }=AB{}^{\prime }$
(d)$\sim (\sim A⨀\sim B)=(A{}^{\prime }⨀B{}^{\prime })=(A{}^{\prime }+B{}^{\prime }){}^{\prime }$
= A $\cdot$ B = A $\wedge$ B
$\therefore$ Only , choice (d) $\equiv A\wedge B$
Note : This problem can also be done by constructing truth table for each choice and comparing with truth table for A $\wedge$ B .