Which one of the following is NOT a valid identity?

Answer: The equality (x + y) ⊕ z = x ⊕ (y + z) is NOT a valid identity.

Counterexample (boolean values 0/1): set x = 0, y = 1, z = 1.

Compute the left side: (x + y) ⊕ z = (0 + 1) ⊕ 1 = 1 ⊕ 1 = 0.

Compute the right side: x ⊕ (y + z) = 0 ⊕ (1 + 1) = 0 ⊕ 1 = 1.

Since the left side yields 0 and the right side yields 1 for this input, the equality fails and therefore is not a valid identity.

Why the other expressions are valid:

• (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z): XOR is associative, so grouping does not affect x ⊕ y ⊕ z (it represents the parity of ones).

• x ⊕ y = x + y, if xy = 0: When x and y are not both 1, OR and XOR give the same result. Check cases: (0,0) → 0, (1,0) → 1, (0,1) → 1.

• x ⊕ y = (xy + x'y')': The expression xy + x'y' is true exactly when x = y (this is XNOR). Its complement is XOR, so the equality holds for all boolean inputs.