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

A Boolean equation is a valid identity only if both sides match for every possible 0/1 assignment of its variables, not just for one convenient case. The standard way to confirm this is to reduce each expression using the core laws — Complement Law (A + Ā = 1), Domination Law (1 + A = 1), Identity Law (A·1 = A), Distributive Law, Absorption Law (A + Ā·B = A + B), and De Morgan's Law — until it settles to a constant or matches the other side; if a reduction does not settle, substituting a concrete assignment that breaks the equality is enough to disprove it.

Checking each of the four expressions against this:

1. ~x + (y·~z) + x + ~y = 1

   By the Complement Law, ~x + x = 1. Substituting leaves 1 + (y·~z) + ~y, and by the Domination Law (1 + A = 1) this stays 1 regardless of y and z.

   Holds for every input → valid identity.

2. x·~y + ~x·y + ~x·~y = 1

   Factor ~x out of the last two terms using the Distributive Law: ~x·y + ~x·~y = ~x·(y + ~y). By the Complement Law, y + ~y = 1, and by the Identity Law, ~x·1 = ~x, so the expression becomes x·~y + ~x.

   This is not a constant — for x = 1, y = 1 it evaluates to (1·0) + 0 = 0, not 1.

   Fails for at least one input → not a valid identity.

3. ~(x + y + z) = ~x·~y·~z

   This is De Morgan's Law itself: the complement of an OR of terms equals the AND of each term's complement.

   Holds for every input → valid identity.

4. x·~y + ~x·y + x·y = x + y

   Group the last two terms: ~x·y + x·y = y·(~x + x) = y. The left side becomes x·~y + y, and by the Absorption Law (A + Ā·B = A + B) this equals x + y — exactly the right side.

   Holds for every input → valid identity.

Cross-check: a second substitution confirms the flagged expression is inconsistent rather than a one-off arithmetic slip. At x = 0, y = 1: x·~y + ~x·y + ~x·~y = (0·0) + (1·1) + (1·0) = 1, but at x = 1, y = 1 it was 0 — the value changes with the input, so it can never be identically 1. By contrast, spot-checking the De Morgan expression at x = 0, y = 1, z = 0 gives ~(0+1+0) = ~1 = 0 on the left and ~0·~1·~0 = 1·0·1 = 0 on the right — consistent, as expected of a genuine identity.

So x·~y + ~x·y + ~x·~y is the one expression among the four that does not hold as a Boolean identity — it is the expression that is not valid.