Question: I have the following non linear programming to solve:

$$\left\{\begin{matrix}
 \min & (x-y)^2 +e^z+e^{-z} \\
 \text{s.t.} & xz=0 \\
 & yz=0 
\end{matrix}\right.$$

The book suggests to make one or more change of variables. I couldn't find a change of variables that makes the non linear programming easier to solve.

I would be glad if someone could help me.

Answer: 
Let $A$ be a statement "Susan goes to school" and $B$ be a stetement "Susan does not talk on the telephone". Then statement "Susan goes to school or Susan does not talk on the phone or Susan does not go to school" may be represented as $A\lor B\lor\neg A$ where $\lor$ is "or" and $\neg$ is "not". Now we have
$$
A\lor B\lor\neg A \equiv A\lor\neg A\lor B \equiv 1\lor B \equiv 1,
$$
where $1$ is truth. So your statement is equvivalent to logical truth, i.e. this statement is always true. 

Indeed, regerdless of whether Susan talks on the phone or not, she definetly goes or not goes to school, so whole statement is true.

---

### Why $x\lor \neg x \equiv 1$ and $1 \lor x \equiv 1$

One may also use the truth tale to show this. Let $1$ be a truth and $0$ be a false. Truth table of "or" operator $\vee$ is 
$$
\begin{array}{cc|c}
x & y & x\lor y \\
\hline
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{array}
$$
Here $x$ and $y$ denotes some statements. This table is may be used as definition of logical operator $\vee$. Note that $x\lor y \equiv y\lor x$ (so operator $\vee$ is associative).

So one may see that $x\lor\neg x = 1$. Indeed, according to $\vee$ truth table:
$$
\begin{array}{cc|c}
 x & \neg x & \lor \neg x \\
\hline
1 & 0 & 1 \\
0 & 1 & 1 \\
\end{array}
$$
so $x\lor\neg x = 1$ regardless of whether $x$ is true or false. 

In the same way one may see that $1\lor x = 1$ for bouth $x$ true and false.
$A\text{ or } B \text{ or not }A$.

One can show by using truth tables that "or" is associative, so that the above is the same as

$\Big(A \text{ or not } A\Big) \text{ or } B.$

The statement $A\text{ or not }A$ is true regardless of whether $A$ is true or false. Hence the statement

$$
\Big(A \text{ or not } A\Big) \text{ or } B
$$
must be true because $E \text{ or }F$ is true if either $E$ is true or $F$ is true.

Hence the given statement must always be true. And "always" means it is true regardless of whether Susan talks on the phone or not, and regardless of whether she goes to school or not. "Always" means "in all four cases". "All four cases" means "all four lines of the truth table".
$$
\begin{array}{c|c|c}
& & \text{Susan goes to school} \\
& & \text{or Susan does not talk on the phone} \\
\text{Susan talks on the phone} & \text{Susan goes to school} & \text{or Susan does not go to school.} \\
\hline
T & T & T \\
T & f & T \\
f & T & T \\
f & f & T \\[6pt]
\hline
& & \uparrow \\[6pt]
& & \text{All four} \\
& & \text{are true.}
\end{array}
$$