How to prove $\textbf{P}(A\cap B^{c}) = \textbf{P}(A) - \textbf{P}(A\cap B)$ I want to use $\textbf{P}(A\cap B^{c}) = \textbf{P}(A) - \textbf{P}(A\cap B)$ as a property in a proof but I'm unsure whether I can use it straight away without proving it. I am unsure how to prove it.

Notice that $A = A\cap\Omega = A\cap(B\cup B^{c}) = (A\cap B)\cup(A\cap B^{c})$, where 
$$(A\cap B)\cap(A\cap B^{c}) = \varnothing$$
Consequently, it results that
\begin{align*}
\textbf{P}(A) & = \textbf{P}((A\cap B)\cup(A\cap B^{c})) = \textbf{P}(A\cap B) + \textbf{P}(A\cap B^{c}) 
\end{align*}
$\therefore \textbf{P}(A\cap B^{c}) = \textbf{P}(A) - \textbf{P}(A\cap B)$, just as desired.
