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

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

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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

3 Answers 3

up vote 0 down vote accepted

Yes, here is the truth table:

enter image description here

(Build by the java applet here)

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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 $x,y,z,t$.)

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

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

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

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

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

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

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

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

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

$(y \Rightarrow t)\lor z$

$\top \lor z$

$\top$

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