GATE CS 2015 Set 3 — Question 16
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Digital Logic → Boolean Algebra & Logic Gates → Boolean Laws & Theorems
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Question
Let # be a binary operator defined aswhere X and Y are Boolean variables.Consider the following two statements.(S1)
(S2)
Which of the following is/are true for the Boolean variables P, Q and R?
(S2)
Which of the following is/are true for the Boolean variables P, Q and R?
Correct answer
(B) Only S2 is true
Solution
The binary operator # is defined as .
Using De Morgan's laws, we can write this as . This is the NAND operation.Statement (S2): Commutativity
We need to check if the operator # is commutative, i.e., if .LHS = .
RHS = .Since Boolean OR (+) is a commutative operation, .
Therefore, the operator # is commutative.
Statement (S2) is true.Statement (S1): Associativity
We need to check if the operator # is associative, i.e., if .LHS =
Using De Morgan's law on , we get .
So, LHS = .RHS =
Using De Morgan's law on , we get .
So, RHS = .Now we check if for all P, Q, R.
Let's test with a counter-example. Let P=1, Q=1, R=0.
LHS = .
RHS = .
Since LHS RHS, the equality does not hold in general. The operator # is not associative.
Statement (S1) is false.Since only statement (S2) is true, the correct option is (B).
Using De Morgan's laws, we can write this as . This is the NAND operation.Statement (S2): Commutativity
We need to check if the operator # is commutative, i.e., if .LHS = .
RHS = .Since Boolean OR (+) is a commutative operation, .
Therefore, the operator # is commutative.
Statement (S2) is true.Statement (S1): Associativity
We need to check if the operator # is associative, i.e., if .LHS =
Using De Morgan's law on , we get .
So, LHS = .RHS =
Using De Morgan's law on , we get .
So, RHS = .Now we check if for all P, Q, R.
Let's test with a counter-example. Let P=1, Q=1, R=0.
LHS = .
RHS = .
Since LHS RHS, the equality does not hold in general. The operator # is not associative.
Statement (S1) is false.Since only statement (S2) is true, the correct option is (B).
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