Which of the following is not a valid Boolean algebra rule?

Concept

A Boolean identity is valid only when its two sides evaluate to the same bit for every possible assignment of its variables.

Two reliable checks are to simplify with standard identities or to find a single counterexample; one counterexample is enough to disprove a universal identity.

Application

1. For X·X = X, the two possible inputs give 0·0 = 0 and 1·1 = 1. This is the idempotent law for AND.

2. For (X+Y)·X = X, distribute to obtain X·X+Y·X. Idempotence gives X+XY, and absorption gives X.

3. For X̄+XY = Y, use A′+AB = A′+B with A = X and B = Y. The left side becomes X̄+Y. At X = 0 and Y = 0, the left side is 1 while the right side is 0, so the claimed universal equality fails.

4. For (X+Y)·(X+Z) = X+YZ, expansion gives X·X+XZ+XY+YZ. Idempotence changes X·X to X, and absorption removes XZ and XY beside X, leaving X+YZ.

Cross-check

The assignment X = 0, Y = 0 directly separates the two sides of X̄+XY = Y as 1 and 0. The other expressions reduce by idempotence, absorption, or distributivity without producing a counterexample.

Result

Therefore, X̄+XY = Y is not a valid Boolean algebra rule.