Prove $P(A \cup B) \leq P(A) + P(B)$ The axioms of probability are:$$\text{(i) }0\leq P(A) \leq1 \text{ for each event $A$}\\ \text{(ii) }P(\Omega)=1\text{ and }P(\varnothing)=0\\
\text{(iii) If $A$ and $B$ are disjoint, then }P(A\cup B)=P(A)+P(B)
\\
\text{(iv) If }A_1, A_2, A_3,... \text{is a sequence of pairwise disjoint events then}\\
P\left( \bigcup_{i=1}^{\infty}A_i\right) = \sum_{i=1}^{\infty} P(A_i)$$
I'm supposed to use the fact that $$A \cup B =(A \cap B^c)+(A\cap B)+(A^c\cap B).$$
Thus, we have$$P(A \cup B) =P(A \cap B^c)+P(A\cap B)+P(A^c\cap B)$$
Not too sure where to go from here.
 A: I would start with noticing that
\begin{align*}
\mathbb{P}(A) = \mathbb{P}(A\cap B) + \mathbb{P}(A\cap B^{c}) \Rightarrow \mathbb{P}(A\cap B^{c}) = \mathbb{P}(A) - \mathbb{P}(A\cap B)
\end{align*}
Similarly, we also have that
\begin{align*}
\mathbb{P}(B) = \mathbb{P}(B\cap A) + \mathbb{P}(B\cap A^{c}) \Rightarrow \mathbb{P}(B\cap A^{c}) = \mathbb{P}(B) - \mathbb{P}(B\cap A)
\end{align*}
Finally, we arrive at the desired inequality:
\begin{align*}
\mathbb{P}(A\cup B) & = \mathbb{P}(A\cap B^{c}) + \mathbb{P}(A\cap B) + \mathbb{P}(A^{c}\cap B)\\\\
& = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A\cap B) \leq \mathbb{P}(A) + \mathbb{P}(B)
\end{align*}
and we are done.
Hopefully this helps!
A: $A\cup B=A\cup (B\cap A^c)$ and since $A$ and $B\cap A^c$ are disjoint,  $P(A\cup B)=P(A)+P(B\cap A^c)$
Also,
$A\cup B=B\cup (A\cap B^c)$ and since $B$ and $A\cap B^c$ are disjoint, $P(A\cup B)=P(B)+P(A\cap B^c)$
Adding these gives $$2P(A\cup B)=P(A)+P(B)+P(A\cap B^c)+P(B\cap A^c)$$
Since $(A\cap B^c)\subset A$, $$P(A\cap B^c)\le P(A)$$
Likewise, $$P(B\cap A^c)\le P(B)$$
Therefore, $$2P(A\cup B)\le 2P(A)+2P(B)\implies P(A\cup B)\le P(A)+P(B)$$
