Addition rule for disjoint sets

Step 1 (base case $k=2$): suppose $A_1$ and $A_2$ are disjoint, so $A_1 \cap A_2 = \emptyset$. Every element of $A_1 \cup A_2$ lies in $A_1$ or in $A_2$, and disjointness rules out lying in both. Splitting $A_1 \cup A_2$ into the two disjoint pieces $A_1$ and $A_2$ and counting each piece once therefore gives $|A_1 \cup A_2| = |A_1| + |A_2|$.

Step 2 (induction on $k$): assume the formula already holds for any $k-1$ pairwise disjoint sets, so $|A_1 \cup \cdots \cup A_{k-1}| = |A_1| + \cdots + |A_{k-1}|$. Set $B = A_1 \cup \cdots \cup A_{k-1}$. Because $A_k$ is disjoint from every $A_i$ with $i < k$, it is also disjoint from their union $B$. Applying the base case to $B$ and $A_k$ gives $|B \cup A_k| = |B| + |A_k|$.

Step 3 (combine): substituting the inductive hypothesis for $|B|$ into the last equality gives $|A_1 \cup \cdots \cup A_k| = |A_1| + \cdots + |A_{k-1}| + |A_k|$, which is exactly the addition rule for $k$ sets. Since the base case $k=2$ holds and each step from $k-1$ to $k$ preserves the formula, it holds for every $k \geq 2$ by induction.