$x \vee y \Rightarrow z \vee t$ - logic I show that $x \Rightarrow z$ and $y \Rightarrow t$ are true.
Is $x \vee y \Rightarrow z \vee t$ then true?
 A: Instead of a brute force approach, one can prove the claim.
Assume $(x\Rightarrow z)\land (y\Rightarrow t)$. Since $(x\lor y)\Rightarrow(z\lor t)$ true if and only if whenever $(x\lor y)$ is valuated as true, then also $(z\lor t)$ does.
$x\lor y$ is true if and only if $x$ is true or $y$ is true.


*

*If $x$ is true, then $z$ is true, therefore $t\lor z$ is true.

*If $y$ is true, then $t$ is true, and thus $t\lor z$ is true.


Either way, if $x\lor y$ is true then $z\lor t$ is true as needed.
(Of course this can formalized completely using the $\operatorname{val}$ function and assignment of truth values for $x,y,z,t$.)
A: Yes, here is the truth table:

(Built by the java applet here)
A: Since $A \Rightarrow B \Leftrightarrow \lnot A \lor B $ we may write :
$\lnot(x \lor y) \lor (z \lor t)$
$(\lnot(x \lor y) \lor z)\lor t$
$((\lnot x \land \lnot y)\lor z)\lor t$
$((\lnot x \lor z)\land(\lnot y \lor z))\lor t$
$((x \Rightarrow z)\land(\lnot y\lor z))\lor t$
$(\top \land (\lnot y\lor z))\lor t$
$(\lnot y\lor z)\lor t$
$(\lnot y \lor t) \lor z$
$(y \Rightarrow t)\lor z$
$\top \lor z$
$\top$
