1. So you've got

1. $p\to q$
2. $\neg p\to x$
3. $y\to z$
4. $a\to \neg x$
5. $\neg y\to \neg q$.

You're asked to prove or disprove $z\to a$.

Since all of your assumptions are implications, the only inference rules you need are modus ponens and modus tollens. In this case, you'd have to assume $z$, and try to show $a$. For modus ponens, you'd need to start somewhere with a $z$ on the LHS of an implication. For modus tollens, you'd need to start somewhere with a $\neg z$ on the RHS of an implication. That does not occur anywhere in your assumptions. Therefore, I deduce that you're not going to be able to prove $z\to a$. But how do you disprove it? You have to be able to assign truth values to all of the propositions such that all your assumptions are true, and yet $z\to a$ is false. In order for $z\to a$ to be false, $z=\text{TRUE}$ and $a=\text{FALSE}.$ So now see if you can assign the rest of the truth values of your propositions such that all the assumptions are satisfied. Make sense?

2. Ahhh yes thank you very much!!! I get it all now, excellent explanation!

3. You're welcome. Have a good one!

4. Originally Posted by Plato
The bit above in red is incorrect.
It should be: $\text{“A only if B”} \equiv \left( {A \to B} \right)$.

Reason: It says that A is true only if B is also true.
The only condition we cannot allow is true implies false.
Thanks for the help guys!

Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?

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5. Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?
See posts 8 - 10.

6. Um why did that person post exactly the same thing as I did on page 1? lol...

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