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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!

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    What does it mean for an implication to be true? How could you write your statement without using implications? – 2017-02-14
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    Generally speaking, say you want to show $A \Rightarrow B \Rightarrow C$, then you simply show $A \Rightarrow B$ then show $B \Rightarrow C$. – 2017-02-14
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    Ahh I see. $(s \Longrightarrow u) = (-s \vee u)$. I could do this to remove every implication. Or should I assume that s is true, so u is true as well. – 2017-02-14
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    @hobomath No, you assume $s \Rightarrow u$. You don't assume $s$, since you are not trying to *show* $s \Rightarrow ...$. – 2017-02-14
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    Alright, I think I work with that! Thanks for your help! – 2017-02-14

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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'.