GATE ECE 2022 | Question: 39

Question

Consider a Boolean gate $\text{(D)}$ where the output $Y$ is related to the inputs $A$ and $B$ as, $Y = A + \overline{B},$ where $+$ denotes logical $\text{OR}$ operation. The Boolean inputs $’0’$ and $’1’$ are also available separately. Using instances of only $\text{D}$ gates and inputs $’0’$ and $’1’,$ _______________ (select the correct option(s)).

• A. $\text{NAND}$ logic can be implemented
• B. $\text{OR}$ logic cannot be implemented
• C. $\text{NOR}$ logic can be implemented
• D. $\text{AND}$ logic cannot be implemented

Answer 1

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https://youtu.be/MJgwNj8y3tw?t=1985&feature=shared

The given boolean gate $D$ is the implication gate. 

Implication gate is functionally complete (with the help of boolean input $0$). Proof HERE.

So, {$D,0$} is functionally complete, hence, we can realize EVERY switching function using these.

So, the answer is $A,C.$

Implementing NOT gate $\bar{B}$ using {$D,0$}: Make input $A = 0.$

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This question (Implication gate) has come in GATE 3 Times:

https://gateoverflow.in/1696/gate-cse-1998-question-5

https://ec.gateoverflow.in/762/gate-ece-2015-set-3-question-37

https://ec.gateoverflow.in/1665/gate-ece-2022-question-39