Hi everyone,

I really need help to prove this statement:

(p→s) from( p→not q), (r→q )and (not r→s)

Thanks

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- Feb 20th 2016, 08:58 AMepsIndirect proof
Hi everyone,

I really need help to prove this statement:

(p→s) from( p→not q), (r→q )and (not r→s)

Thanks - Feb 20th 2016, 09:13 AMArchieRe: Indirect proof
(r→q ) is equivalent to (not q→not r ). The latter is the contrapositive.

- Feb 20th 2016, 09:40 AMPlatoRe: Indirect proof
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 $p\to s~?$ - Feb 20th 2016, 10:47 AMepsRe: Indirect proof
Thank Plato, actually the third promise is (¬r→s) but you wrote ¬(r→s).

- Feb 20th 2016, 11:28 AMPlatoRe: Indirect proof
- Feb 21st 2016, 05:07 AMHallsofIvyRe: Indirect proof
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".