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I need a little help. how can I prove this. This is what I came up with from the question below.
Hypotheses: $p∧q→r, ¬p→s; ¬q→t, ¬r, u→¬s∧¬t$ Conclusion: $¬u$

$P$ = I am able to become rich, $Q$ = I’m willing to become rich, $R$ = I become rich, $S$ = I’m sad, $T$ = I’m lazy, $U$ = I exist.

it reads as such If I were able and willing to become rich, I would do so. If I were unable to become rich, I would be sad; if I were unwilling to become rich, I would be lazy. I do not become rich. If I exist, Iam neither sad nor lazy. Therefore, I do not exist.

Original text :
Problems:
(1) Convert the following argument in logical notation.

If I were able and willing to become rich, I would do so.
If I were unable to become rich, I would be sad;
if I were unwilling to become rich, I would be lazy.
I do not become rich.
If I exist, I am neither sad nor lazy.

Therefore, I do not exist.

Problem 2.
After finishing problem1, check whether the argument is a valid argument. You can do so, by applying various known arguments, such as “modus ponens”, “modus tollens”, “hypothetical syllogism”, “resolution” etc.

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closed as off-topic by Sahiba Arora, Daniel W. Farlow, Leucippus, José Carlos Santos, Glorfindel Jul 31 '17 at 8:07

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  • 4
    $\begingroup$ Please add what your thoughts are about the question. As it stands the question doesn't fit the rules described in the help center. $\endgroup$ – Sahiba Arora Jul 30 '17 at 22:47
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    $\begingroup$ Oops, looks like you left out your own work on this! You can fix that by editing your post to include it. $\endgroup$ – Namaste Jul 30 '17 at 22:58
  • $\begingroup$ I don't really understand what to do after that, not even sure if I came up with the correct premises. $\endgroup$ – cyclist Jul 30 '17 at 23:01
  • $\begingroup$ @cyclist You could start by just listing the rules of inference that you were provided with, and then ask yourself: given what I have, which rules seem applicable? And you can also ask: given what I want, which rule would seem to get me there? Why don't you do that in your post, and then we can provide some more feedback. And please don't immediately give up on symbolic logic: almost everyone is struggling with this at the beginning! You already made the effort to reach out to this community, so I would encourage you to follow up! $\endgroup$ – Bram28 Jul 30 '17 at 23:31
  • $\begingroup$ @amWhy Incorrect. I was trying to help the OP, who seems somewhat downtrodden. $\endgroup$ – Bram28 Jul 31 '17 at 1:19
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Looking at your set of hypothesis, the only one that allows us to conclude $\neg u$ is $\neg(\neg s \wedge \neg t)\to \neg u$, which by De Morgan's laws is equivalent to:

$$s\vee t \to \neg u$$

So, we are done if we can show $s\vee t$. Now, we have $p\wedge q\to r \equiv \neg r \to \neg(p\wedge q)$, and since we also have $\neg r$, we can infer $\neg (p\wedge q)$, i.e. $\neg p \vee \neg q$. Now split it into two cases.

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