What condition of $L(x)$ and $L(y)$ implies $T(x)=T(y)$?

Question

I have the following boolean expression $e_1\land e_2\Leftrightarrow 1$:

$$ \begin{align} \large e_1&:=(\exists t\ge T(y):(t=T(x)\implies L(x)))\\ \large e_2&:=(\exists t\ge T(x):(t=T(y)\implies L(y))) \end{align} $$

Where $t\in\mathbb{R},\enspace f:E\mapsto \mathbb{R},\enspace E=\{e\mid\text{$e$ is an event}\}$ and $L:E\mapsto \{0,1\}$

My attempt:

For $L(x)=L(y)=1$:

Suppose $L(x)\land L(y)=1$, then both are $1$, so the first expression gets simplified to

$$\large\exists t\ge T(y):t=T(x)\Longleftrightarrow T(y)\leq T(x)$$

and the second one to

$$\large T(x)\leq T(y)$$

And the only case where both of them are true is when $T(x) = T(y)$.

---

But what if $L(x)$ or $L(y)$ is false? How can I go around that?

Answer 1

It is a tautology.

If $L(x)\land L(y)\implies T(x)=T(y)$ is true if $L(x)\land L(y)$ is true, then is true too if $L(x)\land L(y)=0$.