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.


1 Answer 1


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

  • $\begingroup$ is there any problem to write $x+z \leq 1$ as $x+z = 1$? $\endgroup$
    – A.Omidi
    Nov 26, 2022 at 9:44
  • 1
    $\begingroup$ @A.Omidi That is too strong. For example, it would mistakenly cut off $(x,y,z)=(0,0,0)$. $\endgroup$
    – RobPratt
    Nov 26, 2022 at 14:31
  • $\begingroup$ many thanks. 🙏 $\endgroup$
    – A.Omidi
    Nov 26, 2022 at 16:03
  • $\begingroup$ The conjunctive normal form derivation is a powerful and general approach, but it can be obscure in simple cases. Here, the correctness of the linear reformulation is clearer if you write it as $z\leq (1-x)$ and $z\leq y$. $\endgroup$
    – 4er
    Apr 2 at 15:24
  • $\begingroup$ @4er Indeed, your rewrite arises immediately from the contrapositive $z \implies (\lnot x \land y)$. $\endgroup$
    – RobPratt
    Apr 2 at 16:15

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