Question

Rules of Infrence help
http://prnt.sc/d4g86p

Answer 1

the commas here can be read as "and" ?

Answer 2

The commas are separating the different hypotheses

Answer 3

Oh, I see.

Answer 4

So overall, there are 4 hypotheses

Answer 5

In general, \(\displaystyle\color{black}{x\to y} \) ≡ \(\displaystyle\color{black}{\neg x\vee y} \). Thus \(\color{black}{ r\to\neg s}\) ≡ \(\displaystyle\color{black}{\neg r\vee \neg s} \).

Then, we have, \(\color{black}{ \neg q\to(r\to\neg s)}\) ≡ \(\displaystyle\color{black}{\neg q\to(\neg r\vee \neg s)} \)
≡ \(\displaystyle\color{black}{q\vee (\neg r\vee \neg s)} \) ≡ \(\displaystyle\color{black}{q\vee \neg r\vee \neg s} \).

Answer 6

So, if we assume that all of the previous steps before conclusion are true, and then \(\displaystyle\color{black}{q\vee \neg r\vee \neg s} \) is also true .... then the conclusion \(\neg s\) will also always be true.

Answer 7

\(\color{black}{ \neg p\to r}\) ≡ \(\displaystyle\color{black}{p\vee r} \).
Then, \(\displaystyle\color{black}{ \neg s} \) can be easily false without making any of the premises not true.

\(\color{black}{ \neg p\to r}\) is true,
\(\displaystyle\color{black}{p\vee r} \) is true
and \(\displaystyle\color{black}{\neg s} \) can still be (possibly) false.

Answer 8

So, if your (Boolean) hypothesis doesn't contain \(s\) (or \(\neg s\)), then it wouldn't be valid to conclude \(\neg s\), because, as again, you can have premises be true and the conclusion be false.

Answer 9

Oh, alright. Because I was trying all methods and only going around in circles. Thank you!