1. ## propositional logic

Hi, Can someone take a look if I have worked through this correctly? thanks

Question 4.
~r -> p, (~p) v q, ~s -> (~p)^(~r), (~p) ^ r -> (~s) v t, ~q, therefore t.

1.~q premise
~q vp premise
therefore ~p disjunctive syllogism

2.~p from 1
~r -> p premise
therefore r modus tollens

3.~p from 1
r from 2

4.~p^r from 3
~s -> (~p)^(~r) premise
therefore s modus tollens

5.~p^r from 3
s from 4
(~p) ^ r -> (~s) v t therefore t disjunctive syllogism

thanks

2. (4) is an incorrect application of modus tollens.

From ~s->(~p^~r), to get s by modus tollens, you need to have ~(~p^~r). Which rules of inference are you using? Can you use De Morgan's Law? Also, in (1), I think you mean, ~p v q, instead of ~q v p. And finally, in step 5, you seem to be compressing to steps into one.

3. I don't know about others, but a lot of your OP is gibberish in my browser. Any chance you could re-type it using LaTeX? You can use \neg for negation, \vee for or, \land for and, \to for implication, etc.

4. Thanks quiney

So in 4
can I use -p^r as a premise?
-p^r is same as -(-rvp) DeMorgans

5. -rvp, -s --> -(-rvp) therefore s Modus tollens

???

5. Since you already have r, try using disjunctive addition to get (pvr). Then use De Morgan's and see what you can get from there.

6. $\neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q$

( $\neg \vee \land \to$) <- my reference**

1. $\neg q$ premise
$\neg p \vee q$ premise
$\therefore \neg p$ disjunctive syllogism

2. $\neg p$ from 1
$\neg r \to p$ premise
$\therefore r$ modus tollens

3. $\neg p$ from 1
$r$ from 2
$\therefore \neg p \land r$ conjunctive addition

4.r from 2
$\therefore p \vee r$

5. $p \vee r$ from 3
I don't know how to use DeMorgans here to get $(\neg p) \land (\neg r)$
I understand DeMorgans to be $(\neg p) \land (\neg r) \equiv \neg (p \vee q)$

??

7. Thanks much for the clarification. It reads much better now.

1. Your step 5 should reference step 4, not step 3.
2. I'm confused as to what exactly your premisses are, and what your desired conclusion is. If you're trying to show $\neg p\land\neg r$ from the following premisses:

$\neg r\to p$
$\neg p\vee q$
$\neg q$,

then I think your proof has just managed to disprove it. Why? Because you've proven $\neg p$, as well as $r$. Therefore, $\neg p\land \neg r$ is false.

8. Originally Posted by dunsta

5. $p \vee r$ from 3
I don't know how to use DeMorgans here to get $(\neg p) \land (\neg r)$
I understand DeMorgans to be $(\neg p) \land (\neg r) \equiv \neg (p \vee q)$
You're not trying to get ~p & ~r, you're trying to get ~(~p & ~r). (p v r) is equivalent to ~(~p & ~r). Remember, De Morgan's works both ways.

Then, use modus tollens and your third premise and you're almost done.

9. $\neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q, \therefore t$

I'm trying to prove t

I can see from here $(\neg p) \land r \to (\neg s) \vee t, \neg q,$ that $(\neg s) \vee t$ means s is true, then -s or t, results in t is true i.e. proving the conclusion.

I too am confused as to how to get there using the basic inference rules, Ackbeet.

I think step 5 needs to show that s is true using $\neg s \to (\neg p) \land (\neg r)$ then I can go on to use the premise from 3 with $(\neg p) \land r \to (\neg s) \vee t$ to show t,

which is what Im working on now with

5. $p \vee r$from 4
$p \vee r \equiv \neg(\neg p \land \neg r)$ DeMorgans

6. $\neg(\neg p \land \neg r)$ from 5
$\neg s \to (\neg p) \land (\neg r)$
$\therefore s$ modus tollens

7. $\neg(p) \land r$ from 3
s from 6
$\neg s \vee t$ premise
$therefore t$ disjunctive syllogism

??

10. $\neg r \to p, (\neg p) \vee q, \neg s \to (\neg p) \land (\neg r), (\neg p) \land r \to (\neg s) \vee t, \neg q, \therefore t$

I'm trying to prove t

I can see from here $(\neg p) \land r \to (\neg s) \vee t$ that $(\neg s) \vee t$ means s is true, then -s or t, results in t is true i.e. proving the conclusion.

I too am confused as to how to get there using the basic inference rules, Ackbeet.

I think step 5 needs to show that s is true using $\neg s \to (\neg p) \land (\neg r)$ then I can go on to use the premise from 3 with $(\neg p) \land r \to (\neg s) \vee t$ to show t,

which is what Im working on now with

5. $p \vee r$from 4
$p \vee r \equiv \neg(\neg p \land \neg r)$ DeMorgans

6. $\neg(\neg p \land \neg r)$ from 5
$\neg s \to (\neg p) \land (\neg r)$
$\therefore s$ modus tollens

7. $\neg(p) \land r$ from 3
s from 6
$\neg s \vee t$ premise
$\therefore t$ disjunctive syllogism

??

11. You've got it. Just clean it up a bit, and, if you wouldn't mind, post the entire proof like this:

Premiss 1
Premiss 2
...
-------------
1. $\neg q$
etc.

n. $t$, from ____.

Make sense?

12. Thanks Ackbeet and quiney, I really mean that.
As it's late here, and I can sleep well now this is solved, I will post the full solution 2mw if you like Ackbeet.

Thanks for taking the time to help.

13. You're welcome for whatever help I was. I do think posting the entire solution will do you good. You'll be able to see better the flow of the whole solution.