Proving (NOT)P given (P v Q ) -> R, R <-> S, (NOT)S

[supasupa]

I want to know if there is a way to prove this without using a truth table... i have begun it but unsure if this is correct...any suggestions would be great


Given that (P v Q ) -> R , R <-> S , (NOT)S are all TRUE. Show that (NOT)P is true.


I have started with (NOT)((P v Q ) V R) which is the same as (P v Q ) -> R but don't see what i can do after this


any suggestions? cheers

 

[loseyourname]

What are the axioms/theorems/inference rules you are allowed to use?

 

[supasupa]

i can use any rules

 

[0rthodontist]

Well, you have to use rules to prove anything. Not knowing what yours are, in general your proof would probably start by showing that R is false. Then you show that (P v Q) is false, and then you show that P is false.

You are incorrect in saying (NOT)((P v Q ) V R) is equivalent to (P v Q) --> R.

 

[loseyourname]

Never mind.

 

[supasupa]

Sorry that was typo... it should read

((NOT)(P v Q ) V R) is equivalent to (P v Q) --> R.

But i don't understand why you wold have to prove that R is false first. What about the other 2 conditions which are also true?

 

[supasupa]

thanks for your help hint... i got it now...

 

[loseyourname]

Present your answer if you got it!

By the way, the reason I asked what rules you could use is that, if you are allowed to use Modus Tollens, you can do it in six steps. If not, but you are allowed to use that equivalency you used (is there a name for that??), then it should take one additional step.

 

[supasupa]

Note: using * for (NOT)
using @ for AND

Given that (P v Q ) -> R , R <-> S , *S are all TRUE. Show that *P is true.

R <-> S is the same as (R->S) @ (S->R)
for this to be true, S if false and hence R is False

If R is false, therefore (P V Q) is False for (P v Q ) -> R to be true
therefore *(*P @ *Q) is False
therefore (*P @ *Q) is true and hence *P and *Q have to be true

 

[neurocomp2003]

i would have started out with teh statement Not(S) and then brought in S<->R or simply Not(S)&S<->R->not(R)

 

[loseyourname]

While all of that is true, it's not the way a derivation is usually written. The proper form is:

1. Premise
2. Premise
C. Conclusion
3. Premise (derived from other premise using axiom or theorem)
4. Premise (derived from other premise using axiom or theorem)
and so on until you derive the conclusion

If your professor will accept that, though, go for it.

 

[supasupa]

i see what you mean...thank you for that...the form is easier to follow and conclude...