Basic probability: If $p(S| \neg G) = 0$, then $p(S|\neg G \& R)$ with $p(\neg G \& R) > 0 = p(S|\neg G) = 0$.

Question

If $p(S|\neg G) = 0$, then $p(\neg G \& R) > 0 \implies p(S\mid \lnot G\& R)=0$.

Could someone please prove this?

Many thanks in advance.

Answer 1

Use the definition of conditional probability.

$$\def\amp{\mathop{\&}}\begin{align}\mathsf P(S\mid\neg G\amp R) &=\dfrac{\mathsf P(S\amp(\lnot G\amp R))}{\mathsf P(\lnot G\amp R)}&&\text{by definition of conditional probability}\\&=\dfrac{\mathsf P(\lnot G\amp(S\amp R))}{\mathsf P(\lnot G\amp R)}&&\text{using commutivity and associativity of }\amp\\&~~\vdots&&\text{and so on...}\end{align}$$