Are these equations derivations of the other?

Question

It is stated P(A|B) = {P(A and B)}{P(B)}

I would like to know if this is a derived from the Bayes' theorem P(A|B) = { P(A)* P(B|A) }/{P(B) }

Answer 1

No, it is the converse. Bayes' Theorem may be derived from the definition of conditional probability. $$\mathsf P(A\mid B)=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\qquad\text{and}\qquad\mathsf P(B\mid A)=\dfrac{\mathsf P(A\cap B)}{\mathsf P(A)}\\\text{therefore}\\~\\\mathsf P(A\mid B)=\dfrac{\mathsf P(A)\,\mathsf P(B\mid A)}{\mathsf P(B)}$$

However, the factor of $\mathsf P(A\cap B)$ cannot be deduced from Bayes' Theorem.