Help with Discrete math problem

[phoenixwhiterose]

Use rules of inference to show that if For all(P(x)vQ(x)) and For all ((notP(x)^Q(x))-->R(x)) are true, then For all((notR(x))-->P(x)) is also true, where the domains of all the quantifiers are the same.
So far all I have is:
1. For all(P(x)vQ(x)) premise
2. P(c)vQ(c) Universal instantiation line 1
3. For all ((notP(x)^Q(x))-->R(x)) premise
4. (((notP(c))^Q(c))-->R(c)) Universal instantiation line 3

Please help me if you can

 

[JakeD]

Use rules of inference to show that if For all(P(x)vQ(x)) and For all ((notP(x)^Q(x))-->R(x)) are true, then For all((notR(x))-->P(x)) is also true, where the domains of all the quantifiers are the same.
So far all I have is:
1. For all(P(x)vQ(x)) premise
2. P(c)vQ(c) Universal instantiation line 1
3. For all ((notP(x)^Q(x))-->R(x)) premise
4. (((notP(c))^Q(c))-->R(c)) Universal instantiation line 3

Please help me if you can

Take the contrapositive of ((notP(x)^Q(x))-->R(x)), then use De Morgan's laws to compute the negative of (notP(x)^Q(x)), then use (P(x)vQ(x)) to get the conclusion P(x).

 

[phoenixwhiterose]

thank you for the help but that not using rules of inference

 

[JakeD]

thank you for the help but that not using rules of inference

I sketched the necessary overall logic, which I did not see in what you had done. That logic needs to be translated to the required form. Everything I used (contrapositive, De Morgan, etc) is in this introduction to logic and the rules of inference. See the important tautologies at the end of section 4.

I've never seen the rules of inference before, but the logic I sketched is sound. If you have difficulty translating the logic to the rules of inference and say specifically where you are stuck, I may be able to help you work through it.

 

[pka]

It is very difficult to know the exact rules you were given.

Here is the argument using Copi’s notation:

\(\displaystyle \neg \left[ {\left( {\forall x} \right)\left( {\neg R(x) \to P(x)} \right)} \right].\) Conditional Proof.
\(\displaystyle \left( {\exists x} \right)\left[ {R(x) \wedge \neg P(x)} \right]\) Negation.
\(\displaystyle \left[ {R(c) \wedge \neg P(c)} \right]\). EI
\(\displaystyle \left( {\neg P(c) \wedge Q(c)} \right) \to R(c)\). UI
\(\displaystyle \neg R(c) \to \left( {P(c) \vee \neg Q(c)} \right)\). Contrapositive.
\(\displaystyle \neg R(c)\). Simplification.
\(\displaystyle \left( {P(c) \vee \neg Q(c)} \right)\) material implication.
\(\displaystyle {\neg P(c)}\) Simplification.
\(\displaystyle {\neg Q(c)}\) Disjunction.
\(\displaystyle \left( {\neg P(c) \wedge \neg Q(c)} \right)\) Contradiction!

Proved.