Demonstrate: $(s \Longrightarrow u) \Longrightarrow (p \wedge((-(u \wedge t) \wedge s ) \vee (q \wedge r))\Longrightarrow -t\vee q\vee r)$

Question

Demonstrate the following theorem: $(s \Longrightarrow u) \Longrightarrow \Bigg(p \wedge \bigg( \Big(-(u \wedge t) \wedge s \Big) \vee (q \wedge r)\bigg)\Longrightarrow -t\vee q\vee r\Bigg)$

I understand how to demonstrate theorems with "$\Leftrightarrow$". However, in this problem, there aren't any "$\Leftrightarrow$" but many "$\Longrightarrow$".

What's the result I should try to get to solve this problem? I don't even know which part of the expression that I should transform... I'd appreciate any hint or help. Thank you!

Answer 1

For this you can use Conditional Proofs:

To prove $\phi \Rightarrow \psi$ using a Conditional proof, you simply assume that $\phi$ is true, and then try to show that $\psi$ will then have to be true as well.

So in your case:

Assume $s \Rightarrow u$

Now show [what comes to the right of the $\Rightarrow$]

but since that is a $\Rightarrow$ as well, do another Conditional Proof, i.e. Also Assume that

$p \land ((\neg (u \land t) \land s) \lor (q \land r))$

And try to show

$\neg t \lor q \lor r$

And that should not be hard:

from the second assumption it follows that $(\neg (u \land t) \land s) \lor (q \land r)$, and thus you have either $\neg (u \land t) \land s$ or you have $q \land r$, and in either case you can establish your goal $\neg t \lor q \lor r$ (use the initial assumption for the first case)

So you don't really transform any statements here like you did algebraically when working with $\Leftrightarrow$'s. Rather, you infer things from other things ... And thus you get $\Rightarrow$'s, since inference is (typically) a 'one-way street'.