# Simplification of boolean algebra from “not s and p” to “not s”

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:

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

Any help is welcome.

Thanks.

-

$\lnot s \not\equiv (\lnot s \land p),\;$ 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 $p$.
-
So when I have this type of assingments, where I need to rules of inference, I can presume that all of the expressions are true, and do assingment from there. Right? How do I know is it true or not? – depecheSoul Oct 26 '13 at 15:11
@amWhy: Right up your alley! =1 – Amzoti Oct 27 '13 at 0:07

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).

-