GATE CSE 1994 | Question: 4

Question

• A. Let $\ast$ be a Boolean operation defined as $A\ast B = AB + \overline{A}\;\overline{B}$. If $C=A\ast B$ then evaluate and fill in the blanks:
  1. $A\ast A=$______
  2. $C\ast A=$______

• B. Solve the following boolean equations for the values of $A, B$ and $C:$
  $AB+\overline{A}C=1$
  $AC+B=0$

Comments

Key Point: * is ex-nor Boolean operator, and it is commutative as well as associative.

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for part $(B)$, easy way to do is to create a table for all the 8 cases and keep ruling out which doesnot satisfy our conditions. Whichever is left at the end is our answer.

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$\color{red}{\text{Find Detailed Video Solution Below:}}$ 

https://youtu.be/ZYgUWO_omuI

Answer 1

• A.  
  1. $A\ast A = AA+A'A' = A + A' = 1$
  2. $C\ast A= (A\ast B)\ast A = (AB + A'B')\ast A= (AB + A'B')A +(AB + A'B')'A'$
     $\qquad =(AB + A'B')A+(A'B+AB')A' = AB +0 +A'B+0 = B.$

 

• A. $AB + A'C= 1, AC + B = 0$

$\implies AC + B = 0,$ means both $B = 0$  and $AC = 0$

$\implies AB+ A'C = 1$

$\implies A'C = 1 \qquad \left[\because B = 0, AB = 0 \right]$

So, $C = 1$ and $A = 0$

$A = 0 , B = 0$ and $C = 1$

Comments

Sir , plz explain this part  once more 

AB + A'C= 1 , AC + B = 0

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AC+ B , means OR, OR is 0 only when both inputs in OR are 0

means B is 0 and AC is 0

AB+AC'= 1

as B is 0 already known now, then AB is also 0.

OR is 1 , when atleast one of input is 1, but AB is 0 so A'C should be 1

now A'C =1 , this is AND of A' and C. AND is 1 , when both inputs are 1

mean A' =1 and C =1

A'=1 , means A = 0

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PART a. :

here boolean operation * denotes the equivalence function $\color{RED}{⊙}$.
It evaluates to true only when both the variables have same values.
$F(x,y)=x\odot y= \left\{\begin{matrix} 1 & x=y=1 \ or \ x=y=0 \\ 0&Otherwise \end{matrix}\right.$

So,
$i) A* A  \rightarrow always \ 1$ 
$A=0 , \ \  \ \  0*0=1$ 
$A=1 , \ \ \ \  1*1=1$

$ii) C∗A$
$A∗B*A = A*A*B = 1*B = B$       (equivalence operation is both commutative and associative) 
$B=0 , \ \  \ \  1*0=0$ 
$B=1 , \ \ \ \  1*1=1$

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• $\textbf{EX-OR}(\oplus)$ and $\textbf{EX-NOR}(\odot)$ is  both commutative and associative.
• $\textbf{NAND} $ and $\textbf{NOR}$ is both commutative but both is not associative.

References:

• https://math.stackexchange.com/questions/293793/prove-xor-is-commutative-and-associative
• https://www.mi.mun.ca/users/cchaulk/misc/boolean.htm

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the B part of the answer is just lit