Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I show that $x \Rightarrow z$ and $y \Rightarrow t$ are true.

Is $x \vee y \Rightarrow z \vee t$ then true?

share|cite|improve this question
I see that the answers here use truth tables or otherwise implicitly use the law of the excluded middle. This is more general than that though, and still holds in constructive logic. – George Lowther Oct 3 '11 at 11:16
up vote 0 down vote accepted

Yes, here is the truth table:

enter image description here

(Built by the java applet here)

share|cite|improve this answer

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 $x,y,z,t$.)

share|cite|improve this answer

Since $A \Rightarrow B \Leftrightarrow \lnot A \lor B $ we may write :

$\lnot(x \lor y) \lor (z \lor t)$

$(\lnot(x \lor y) \lor z)\lor t$

$((\lnot x \land \lnot y)\lor z)\lor t$

$((\lnot x \lor z)\land(\lnot y \lor z))\lor t$

$((x \Rightarrow z)\land(\lnot y\lor z))\lor t$

$(\top \land (\lnot y\lor z))\lor t$

$(\lnot y\lor z)\lor t$

$(\lnot y \lor t) \lor z$

$(y \Rightarrow t)\lor z$

$\top \lor z$


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.