Simplification of boolean algebra from "not s and p" to "not s"

Question

I am trying to learn more about "Rules of Inference" and their application, but one thing always confuses me, and that is simplification "not s and p" to "not s".

I have looked at some examples:

http://www.site.uottawa.ca/~lucia/courses/2101-10/lecturenotes/04InferenceRulesProofMethods.pdf page 18

http://www2.cs.siu.edu/~nojoumian/CS215/Files/Lec06_CS215.pdf page 16

And I simply dont understand how is it possible to reduce expression.

Any help is welcome.

Thanks.

Answer 1

$\lnot s \not\equiv (\lnot s \land p),\;$ but it is the case that $\;\lnot s\;$ follows from $\;\lnot s \land p$.

"$\lnot s \land p$ is true" means

• $\lnot s$ is true, and $ p $ is true.

So it certainly follows that

• "$\lnot s$ is true,"

just as it follows that

• $p$ is true.

More simply put, we have

• $\lnot s$ AND $p$.
  • Therefore $\lnot s$.
  • Therefore $p$.

Answer 2

This simplification is not intended to be in the sense of rewriting a nicer-looking, equivalent statement. Instead, the simplification is a nicer-looking statement that logically follows from the first. Of course, if $\neg s\vee p$ is true, then $\neg s$ is true (and also $p$ is true, so $p$ would be another simplification in this sense).