1. ## Indirect proof

Hi everyone,
I really need help to prove this statement:
(p→s) from( p→not q), (r→q )and (not r→s)

Thanks

2. ## Re: Indirect proof

(r→q ) is equivalent to (not q→not r ). The latter is the contrapositive.

3. ## Re: Indirect proof

Originally Posted by eps
Hi eve
need help to prove this statement: (p→s) from( p→not q), (r→q )and (not r→s)
Here are the premises: $I~(p\to\neg q)~,~II~(r\to q)~\&~III,~\neg(r\to s).$

To prove by indirect proof you must assume that $A.~(p\wedge \neg s)$ is true.
From A. & I. you can get $\neg q$. HOW?
Then from II. you get $\neg r$
Then from III. you get $r$
So how has it been proven that $p\to s~?$

4. ## Re: Indirect proof

Thank Plato, actually the third promise is (¬r→s) but you wrote ¬(r→s).

5. ## Re: Indirect proof

Originally Posted by eps
Thank Plato, actually the third promise is (¬r→s) but you wrote ¬(r→s).
But you said not r implies s. The usual order of operations would be (not r) implies s if that is what you meant.
It still works. from $\neg r$ and $\neg r\to s$ you get $s$ modus pomens.
Then from that and A. you get $s\wedge\neg s$ DONE>

6. ## Re: Indirect proof

Originally Posted by eps
Hi everyone,
I really need help to prove this statement:
(p→s) from( p→not q), (r→q )and (not r→s)

Thanks
To continue from " p→not q" my next implication would have to start "not q". So I recognize that r→q is equivalent to it contrapositive, "not q→ not r". Replacing "r→ q" with "not q→ not r", I have the chain
"p→ not q", "not q→ not r" and "not r→ s" from which I immediately deduce "p→ s".