Different versions of Bayes Theorem

Question

Given this formula:

$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{P(A∧B)}{P(A∧B)+P(¬A∧B)}$$

The second part is the Bayesian formula, the third part is what I thought the Bayesian formula must be based on my own reasoning.

Question:

1. Is this equality true? (Is my reasoning correct?)
2. If they are the same; then how can I transform one into the other?
3. More in general: where can I find rules for working with these kinds of functions?

Answer 1

$$P(A|B)=\frac{P(A∧B)}{P(A∧B)+P(¬A∧B)}$$

This is true, and is essentially the definition of $P(A|B)$. Note however that the denominator $P(A∧B)+P(¬A∧B)$ is equal to $P(B)$, so we have:

$$P(A|B)=\frac{P(A∧B)}{P(B)}$$

From this definition, you can deduce Bayes. Just write the same formula with A and B swapped:

$$P(B|A)=\frac{P(B∧A)}{P(A)}$$

Then divide the last two equations, and use the fact that $A∧B=B∧A$, you get Bayes: $$\frac{P(A|B)}{P(B|A)}=\frac{P(A∧B)}{P(B)}\frac{P(A)}{P(B∧A)}=\frac{P(A)}{P(B)}$$