The problem you have provided asks to prove the inequality related to the union and cardinality of two finite sets $A$ and $B$:

$\#(A \cup B) \leq \#A + \#B$ and the equality holds if and only if $A \cap B = \emptyset$ (i.e., $A$ and $B$ are disjoint).

Proof Outline:

1. Cardinality of the Union Formula:
   There is a well-known formula for the cardinality of the union of two finite sets: $\#(A \cup B) = \#A + \#B - \#(A \cap B)$ This formula counts the elements of $A$ and $B$, but since elements in $A \cap B$ are counted twice (once in $A$ and once in $B$), we subtract $\#(A \cap B)$ to avoid overcounting.

2. Inequality Condition:
   From the formula above, we have: $\#(A \cup B) = \#A + \#B - \#(A \cap B)$ Since the cardinality of $A \cap B$ is non-negative (i.e., $\#(A \cap B) \geq 0$), we get: $\#(A \cup B) \leq \#A + \#B$ This proves the inequality.

3. Equality Condition:
   The equality holds if and only if $\#(A \cap B) = 0$, which means that $A$ and $B$ are disjoint sets (i.e., $A \cap B = \emptyset$).

Thus, we have proven that: $\#(A \cup B) \leq \#A + \#B$ with equality holding if and only if $A \cap B = \emptyset$.

Key Set Operations Used: