Prove ~s=>~p given (r=>s) and (p|q)=>(r|s)

Question

I am trying to prove ~s=>~p (not s implies not p) given the following 2 premises.

   r=>s          [r implies s] 

   (p|q)=>(r|s)  [(p or q) implies (r or s)]

I have tried several ways, trying to use OR elimination or Negation Introduction, but I can't even visualize which assumptions I will need to use. Would appreciate any and all help that can be provided.

score 0 · Answer 1 · answered Dec 03 '13 at 06:09

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Maybe you're missing that you can combine the two givens before anything else, to eliminate an r term. I don't think you need negation introduction, contrapositing a statement is sufficient.

answered Dec 03 '13 at 06:09

piojo

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score 0 · Answer 2 · answered Dec 03 '13 at 06:17

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(p|q)=>(r|s)
(p|q)=>(s|s) //r=>s
(p|q)=>s //simplify
~s=>~(p|q) //by contraposition
~s=>~p and ~s=>~q
~s=>~p

answered Dec 03 '13 at 06:17

IceArdor

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I can follow the reasoning above, but the main difficulty I'm having is translating logic/common sense to the Fitch language. I have an online proof outline, but only have the basic 10 reasons available to use. Thanks to everyone for the speedy replies, I appreciate all your assistance. – user3060136 Dec 03 '13 at 16:22

Yick Leung · Answer 3 · 2013-12-03T07:22:38.470

I will prove this by contradiction.

~S=>~P is logically equivalent to P=>S.

P=>S is logically equivalent to ~PvS.

Let v mean "or" and & mean "and".

Suppose ~PvS is false.

Therefore, ~(~PvS) is true. (This just means that the negation of it will be true.)

~(~PvS) = P&~S (De Morgan's Law) -----------(1)

So, if our assumption is correct, then all three statements that we have: P&~S, R=>S, and (PvQ)=>(RvS) should be all true.

(PvQ)=>(RvS) is logically equivalent to ~(PvQ)v(RvS). Which is equivalent to (~P&~Q)v(RvS).-------------------(2)

The other premise R=>S is equivalent to ~RvS. ----------(3)

If (1) is true from out assumption, then both P and ~S have to be true. This is because of the nature of the & logical connective. ~S is true, so S must be false. Now we substitute P=True and S=False into (2).

On the Left hand side: If P is true, then ~P must be false. Because of the nature of the & connective, (~P&~Q) must be false regardless of what ~Q is.

So now the Right hand side: (RvS) must be true if we need (2) to be true. Since S is false, then R must be true.

We have now deduced that: S is false, R is true, P is true.

Now we can substitute these truth values into (3). Since S is false. Then ~R must be true. Hence, ~(~R) is false. Thus, R is false.

However, contradiction with the fact that R is true. So, our original assumption that ~S=>~P is false was wrong. Therefore, ~S=>~P is true.

At the end of the day, the logical equivalences that were mentioned previously can be verified by using a truth table. But it is good to memorize them. Cheers.

Prove ~s=>~p given (r=>s) and (p|q)=>(r|s)

3 Answers3