To prove the statement: If A B and card(A) = card(B), then A=B. This proof is valid for finite sets.

We are given two finite sets $A$ and $B$ such that:

• $A \subset B$ (This means every element of $A$ is also an element of $B$).
• $\text{card}(A) = \text{card}(B)$ (This means $A$ and $B$ have the same number of elements).

To prove that $A=B$, we must show that $A \subset B$ and $B \subset A$. We are already given $A \subset B$, so we only need to prove $B \subset A$.

Assume, for the sake of contradiction, that $B \not\subset A$.
If $B \not\subset A$, it means there exists at least one element $x$ such that $x \in B$ and $x \notin A$.
Since $A \subset B$, all elements of $A$ are already included in $B$.
If there is an element $x$ in $B$ that is not in $A$, then $B$ contains all the elements of $A$ plus at least one additional element ($x$).
This implies that the number of elements in $B$ must be strictly greater than the number of elements in $A$.
Thus, we would have $\text{card}(B) > \text{card}(A)$.

The conclusion from Step 3, $\text{card}(B) > \text{card}(A)$, directly contradicts our given condition that $\text{card}(A) = \text{card}(B)$.
Therefore, our initial assumption that $B \not\subset A$ must be false.
This means that $B \subset A$ must be true.

Since we have established both $A \subset B$ (given) and $B \subset A$ (proven), we can conclude that:

This completes the proof of the auxiliary hypothesis.