Let # be a binary operator defined as 𝑋 # π‘Œ = 𝑋′ + π‘Œβ€² where 𝑋 and π‘Œ are…

2015

Let # be a binary operator defined as

𝑋 # π‘Œ = 𝑋′ + π‘Œβ€² where 𝑋 and π‘Œ are Boolean variables.

Consider the following two statements.

(S1) (𝑃 # 𝑄)#𝑅 = 𝑃#(𝑄 # 𝑅)

(S2) 𝑄#𝑅 = 𝑅#𝑄

Which of the following is/are true for the Boolean variables 𝑃, 𝑄 and 𝑅?

  1. A.

    Only S1 is true

  2. B.

    Only S2 is true

  3. C.

    Both S1 and S2 are true

  4. D.

    Neither S1 nor S2 are true

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Correct answer: B

Key idea: the operator # is X # Y = X' + Y' which by De Morgan is (X AND Y)'.

  • Check commutativity: Q # R = (Q AND R)' = (R AND Q)' = R # Q, so the commutativity statement is true.

  • Check associativity (use a counterexample): let P = 1, Q = 1, R = 0. Then (P # Q) # R = ((1 AND 1)') # 0 = (1') # 0 = 0 # 0 = (0 AND 0)' = 1, while P # (Q # R) = 1 # ((1 AND 0)') = 1 # (0') = 1 # 1 = (1 AND 1)' = 0. The two results differ, so associativity fails.

Conclusion: The commutativity statement is true and the associativity statement is false, so only the commutativity statement holds.

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