GATE CS 2018 Q4 — Boolean Algebra

To find the incorrect identity, let us analyze each option using the properties of XOR ($\oplus$) and XNOR ($\odot$):

• Option A: $\overline{P \oplus Q} = P \odot Q$
  By definition, XNOR is the complement of XOR. Thus, this identity is correct.
• Option B: $\overline{P} \oplus Q = P \odot Q$
  Complementing one input of an XOR gate complements the entire output, which converts it into an XNOR gate.
  $\overline{P} \oplus Q = \overline{P}\overline{Q} + P Q = P \odot Q$. Thus, this identity is correct.
• Option C: $\overline{P} \oplus \overline{Q} = P \oplus Q$
  Complementing both inputs of an XOR gate leaves the output unchanged.
  $\overline{P} \oplus \overline{Q} = \overline{P}\overline{(\overline{Q})} + \overline{(\overline{P})}\overline{Q} = \overline{P}Q + P\overline{Q} = P \oplus Q$. Thus, this identity is correct.
• Option D: $P \oplus \overline{P} \oplus Q = P \odot \overline{P} \odot \overline{Q}$
  Let's evaluate the Left-Hand Side (LHS) and Right-Hand Side (RHS) separately:
  • $\text{LHS} = (P \oplus \overline{P}) \oplus Q = 1 \oplus Q = \overline{Q}$
  • $\text{RHS} = (P \odot \overline{P}) \odot \overline{Q} = 0 \odot \overline{Q} = \overline{0 \oplus \overline{Q}} = \overline{\overline{Q}} = Q$
  Since $\text{LHS} = \overline{Q} \neq Q = \text{RHS}$, this identity is NOT CORRECT.