Binary variable constraint for condition

Question

I am trying to solve the following task: If $x=1$ or $y=0$ then $z=0$

My approach:

If $z=0$ then $x+y \le 2 + Mz \implies x+y \le 2+2z \quad$ where $M = 2$

If $z=1$ then $x+y=1 \\ \implies x+y \le 1, \quad x+y \ge 1 - M(1-z) \\ \implies x+y \ge z$

where $M = 1$

I already tried to add the constraint: $y-x \ge 0$ to ensure that combination $x=1, y=0, z=1$ should be satisfied and combination $x=1, y=0, z=1$ should be not satisfied.

But then the combination $x=1, y=0, z=0$ isn’t satisfied, although it should be.

In 1 combination the sum of $x+y = 2$.

I would be grateful for any hints regarding the task or showing where my approach is wrong.

Answer 1

Via conjunctive normal form, $$ (x \lor \lnot y) \implies \lnot z \\ \lnot (x \lor \lnot y) \lor \lnot z \\ (\lnot x \land y) \lor \lnot z \\ (\lnot x \lor \lnot z) \land (y \lor \lnot z) \\ ((1 - x) + (1 - z) \ge 1) \land (y + (1 - z) \ge 1) \\ (x + z \le 1) \land (y \ge z) $$