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Boolean Algebra and Minimization

GATE CSE 2019 · Q6

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Time

Which one of the following is NOT a valid identity?

Select one answerSingle correct

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Solution

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Official answer: B) (x + y) ⊕ z = x ⊕ (y + z)

Boolean Algebra and Minimization

Let us examine each option:

  • Option A: XOR is associative, so (x⊕y)⊕z=x⊕(y⊕z)(x \oplus y) \oplus z = x \oplus (y \oplus z) is a valid identity.

  • Option B: (x+y)⊕z=x⊕(y+z)(x + y) \oplus z = x \oplus (y + z) is not valid. For example, let x=1x = 1, y=1y = 1, z=0z = 0:

    LHS=(1+1)⊕0=1⊕0=1\text{LHS} = (1 + 1) \oplus 0 = 1 \oplus 0 = 1
    RHS=1⊕(1+0)=1⊕1=0\text{RHS} = 1 \oplus (1 + 0) = 1 \oplus 1 = 0

    Since LHS≠RHS\text{LHS} \neq \text{RHS}, this is not a valid identity.

  • Option C: x⊕y=x′y+xy′x \oplus y = x^{\prime}y + xy^{\prime}. If xy=0xy = 0, then x+y=x′y+xy′+xy=x′y+xy′=x⊕yx + y = x^{\prime}y + xy^{\prime} + xy = x^{\prime}y + xy^{\prime} = x \oplus y. Thus, this is valid.

  • Option D: xy+x′y′xy + x^{\prime}y^{\prime} is the standard XNOR expression. The complement of XNOR is XOR, so (xy+x′y′)′=x⊕y(xy + x^{\prime}y^{\prime})' = x \oplus y, which is valid.

Therefore, B is NOT a valid identity.

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This Boolean Algebra and Minimization previous year question is Q6 of the GATE CSE 2019 paper. Find more GATE CSE PYQs on Boolean Algebra and Minimization in Digital Logic, or browse all GATE CSE PYQs.

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Boolean Algebra and Minimization in GATE CSE

Boolean Algebra and Minimization has 36 GATE CSE PYQs across 23 papers from 2011 to 2026. 13 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2019.

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Which one of the following is NOT a valid identity?

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.