For a noninductive argument for Proposition 4, note first that if a point of the sample space is not a menber of any of the sets Ei then its probability does not contribute anything to either side of the equality. On the other hand, suppose that a point is in exactly m of the events Ei, where m>0. Then since it is in $\displaystyle\bigcup_i E_i$ its probability is counted once in $\displaystyle P\left (\bigcup_i E_i\right )$; also as this point is contained in $\displaystyle{m\choose k}$ subsets of the type $\displaystyle E_i_1 E_i_2\cdots E_i_k$, its probability is counted

$\displaystyle{m\choose 1}-{m\choose 2}+{m\choose 3}-\cdots\pm{m\choose m}$
times on the right of the equality sign in Proposition 4. Thus, for m>0, we must show that
$\displaystyle1={m\choose 1}-{m\choose 2}+{m\choose 3}-\cdots\pm{m\choose m}$
However, since $\displaystyle1={m\choose 0}$, the preceding is equivalent to
$\displaystyle\sum^m_{i=0} {m\choose i}(-1)^i=0$
and the latter equation follows from the binomial theorem since
$\displaystyle0=(-1+1)^m=\sum^m_{i=0} {m\choose i}(-1)^i(1)^{m-i}$