Prove that $P(E \cap F^c) = P(E) - P(E \cap F)$ I'm trying to prove the equality $P(E \cap F^c) = P(E) - P(E \cap F)$, but I feel as if I'm going in circles after I wrote one of the probability axioms $P(E \cap F^c) = P(E) + P(F^c)-P(E \cup F^c)$.
Observing that we have $P(E)$ on both sides, I tried to prove that $-P(E\cap F) = P(F^c)-P(E\cup F^c)$, or better yet $P(E \cup F^c) = P(F^c)+P(E\cap F)$.
I used the probability axiom once again to achieve the equality $P(E \cup F^c) = P(E) + P(F^c) - P(E \cap F^c)$.
At this point, I feel lost since I think I'm going in circles. I tried to show that $P(E)-P(E\cap F^c) = P(E\cap F)$. However, I think it's only intuitively true when $E$ and $F$ are disjoint, which is not part of the assumption.
I thought this problem might have been easy since I'm pretty comfortable with elementary set theory, but the fact that these are real numbers between 0 and 1 is throwing me off. I'm not yet sure how to think about proving statements about probability, so intuition is especially appreciated.

Split the sets into mutually exclusive sets.
In this case $E = (E \cap F) \cup (E \setminus F)$
and the sets 
$E \cap F$, $E \setminus F$  are pairwise
disjoint.
Hence $P(E) = P(E \cap F) + P(E \setminus F)$.
