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Let A$1$,A$2$,...,An be events, and let N denote the number of them that occur.

Given the values as follow,

$P(A_1鈰�A_n)鈮�\underset{i=1}{\overset{n}{鈭�}}P\left(A_i\right)鈭�(n鈭�1)$

N$鈮�$ n 鈭� 1 + $1$.

$\text{Let}N\text{represents the number of events}{A}_1,A_2,鈥�,A_n\text{that occur and let indicator variable}I\text{be defined as:}$

localid="1648086527186" $I=\left\{\begin{array}{l}1,\text{if}{A}_1{A}_2鈰�A_n\text{occurs}\\ 0,\text{if}{A}_1{A}_2鈰�A_n\text{does not occur}\end{array}\right.$

$\text{Further, if indicator variables}{I}_j,j=1,2,鈥�,n\text{, are defined as:}$

localid="1648086533299" $I_j=\left\{\begin{array}{l}1,\text{if}{A}_j\text{occurs}\\ 0,\text{if}{A}_j\text{does not occur}\end{array}\right.$

Then,

$N=\underset{j=1}{\overset{n}{鈭�}}{I}_j$

If all of the events $A_i,i=1,2,鈥�,n$occur, then$N=nandI=1$

and in that case, variable $N$can be represented as

$N=n鈭�1+1=n鈭�1+I$

On the other hand, if all of the events $A_i,i=1,2,鈥�,n$do not occur, then

$N<n\text{and}I=0$

and in that case

$N鈮�n鈭�1+0=n鈭�1+I.$

$\text{Hence,}N鈮�n鈭�1+I$

Now, use induction to prove Bonferroni's inequality.

The first case is $n=1$, which is the trivial statement, since

$P\left(A_1\right)=P\left(A_1\right)$

Let's take the second case.

In second case it is $n=2$since $1鈮�P\left(A_1鈭�A_2\right)=P\left(A_1\right)+P\left(A_2\right)鈭�P\left(A_1{A}_2\right)$.

Here, we have $P\left(A_1{A}_2\right)鈮�P\left(A_1\right)+P\left(A_2\right)鈭�1$

Therefore, the statement is true.

From the above steps, we obtain that, for events $A_1,A_2,鈥�,A_n$, we have

$P\left(A_1鈰�A_n\right)鈮�\underset{i=1}{\overset{n}{鈭�}}鈥�P\left(A_i\right)鈭�(n鈭�1)$

Let's prove the desired statement for $n+1$.

Therefore, consider the events$A_1,A_2,鈥�,A_n,A_{n+1}$. Because$A_1,A_2,鈥�,A_{n鈭�1},A_n鈭�A_{n+1}$is a list of$n$events.

$P\left(A_1{A}_2鈰�A_n{A}_{n+1}\right)=P\left(A_1鈭�鈰�A_{n鈭�1}鈭�A_n鈭�A_{n+1}\right)$$=P{\left(A_1{A}_2鈰�A_{n鈭�1}\left(A_n鈭�A_{n+1}\right)\right)}^{\text{inductive hypothesis}}鈮�P\left(A_1\right)+鈰�+P\left(A_{n鈭�1}\right)+P\left(A_n鈭�A_{n+1}\right)鈭�(n鈭�1)$

But, since the statement is true for $n=2$, we have that:

$P\left(A_n鈭�A_{n+1}\right)鈮�P\left(A_n\right)+P\left(A_{n+1}\right)鈭�1$

Therefore, we can concluded that

$P\left(A_1{A}_2鈰�A_n{A}_{n+1}\right)鈮�P\left(A_1\right)+鈰�+P\left(A_{n鈭�1}\right)+\left[P\left(A_n\right)+P\left(A_{n+1}\right)鈭�1\right]鈭�(n鈭�1)$

$=P\left(A_1\right)+鈰�+P\left(A_{n鈭�1}\right)+P\left(A_n\right)+P\left(A_{n+1}\right)鈭�n$

$=P\left(A_1\right)+鈰�+P\left(A_{n鈭�1}\right)+P\left(A_n\right)+P\left(A_{n+1}\right)鈭�\left(\right(n+1)鈭�1)$

If all of the events $A_i,i=1,2,鈥�,n$occur, then

$N=n$ and $I=1$,

while on the other side, if all of the events $A_i,i=1,2,鈥�,n$do not occur, then

$N<n\text{and}I=0$

Therefore,$N鈮�n鈭�1+I$