How to prove $P(A\cap B)\geqslant P(A)+P(B)-1$?

I want to prove $P(A\cap B)\geqslant P(A)+P(B)-1$.

How can I simplify the following proof?

$P(A\cap B)=P(A)-P(A\cap B^\mathsf{c})=P(A)-1+P(A^\mathsf{c}\cup B)=P(A)-1+P(A^\mathsf{c})+P(B)-P(A^\mathsf{c}\cap B)=P(A)+P(B)+P(A^\mathsf{c}\cap B^\mathsf{c})-1\geqslant P(A)+P(B)-1$

• You can simplify the "proof" by throwing it away since it is incorrect. In the second equality, you have mistakenly substituted $P(A^c \cup B)$ for $P(A\cap B^c)$. In fact, $$P(A^c \cup B) = 1 - P(A\cap B^c)$$ since DeMorgan's laws assure us that $(A\cap B^c)^c = A^c \cup B$. – Dilip Sarwate Dec 6 '14 at 15:23
• It might help the intuition to note that you are being asked to prove $P(A^c\cup B^c)\le P(A^c) + P(B^c)$. – whuber Dec 6 '14 at 18:31

Just use the inclusion-exclusion formula: $$1 \geq P(A\cup B) = P(A) + P(B) - P(A\cap B),$$ then rearrange terms.