Hi everyone,
I really need help to prove this statement:
(p→s) from( p→not q), (r→q )and (not r→s)
Thanks
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 $p\to s~?$
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".