Prove $\bigcap \{A,B\} = A \cap B$ I am trying to prove theorem 71 of Patrick Suppes Axiomatic Set Theory that ∩{A,B} = A ∩ B. I have a proof but it makes extreme use of propositional logic starting with the definitions of ∩ and {A,B}. Is there another more elegant proof?
 A: Hint
The proof must be :
1) let $x \in A \cap B$; then $x \in A \ $ and $ \ x \in B$.
By Def.14, page 39 : $x \in \cap A \ $ iff $ \ \forall B \ (B \in A \to x \in B)$.
But from $x \in A$ and $x \in B$ we have that $x \in \{ A,B \}$; thus it is true that :

$\forall Z \ (Z \in \{ A,B \} \to x \in Z)$,

because the only two elements of $\{ A,B \}$ are precisely $A$ and $B$; thus, we can apply the definition concluding with : $x \in \cap \{ A,B \}$.
Thus, we have that :

if $x \in A \cap B$, then $x \in \cap \{ A,B \}$.

Being $x$ "generic", we have that the above holds for all $x$, i.e. :

$\forall x \ (x \in A \cap B \to x \in \cap \{ A,B \})$

and by definition of inclusion, we conclude with :

$$A \cap B \subseteq \cap \{ A,B \}.$$


The other "inclusion" is similar :
2) let $x \in \cap \{ A,B \} \ldots$
A: Here is my proof.

Is there anything wrong with it?

Theorem: $\cap\;\{A, B\}=A\cap B$


Proof:
Suppose $\forall Z,z\in \cap\{A,B\}$ is true.
By the definition $13,$ $$\cap\{A,B\}=\{x|(\forall C)(C\in\{A B\}\to x\in C)\}.$$
Therefore, by theorem $41,$
$$(\forall C)\left[(C\in\{A,B\})_{\to} z\in C\right]$$ is true, but because
$$A\in\{A, B\}$$ is true, $z\in A$ is true.
Similarly
because $$B\in\{A,B\}$$ is true, $z\in B$ is true.
Therefore, $(z \in A \bullet z \in B)$ is true.
By the theorem $14,$ $$z \in(A \cap B).$$ Consequently, through $$CP
[(\forall z)(z\in \cap\{A, B\} \to z\in(A\cap B)]$$ has been
proven.
By the definition of subsets $1,$ $$\cap\{A, B\}\subseteq(A\cap B)$$ is true.
Now,suppose $$\forall Z, z\in(\ A\cap B)$$is true.
Therefore, by the theorem $14,$ $$(z\in A\bullet z\in B)$$ is true.
Consequently, $z\in B$ is
true, but $$(\forall B)(B\in\{A,B\})$$ is true s.t.
$$(\exists B)(B\in\{A,B\})$$ is also true.
Therefore,
$$[(\forall B)\in\{A,B\}\to z\in B) \bullet(\exists B)(B\in\{A, B\})]$$ is
true.
Consequently, by the theorem $62$, $$z \in n\{A,B\}$$ is true.
Therefore, by $CP,$ $$[(\forall z)(z\in(A\cap B) \to z\in \cap\{A,B\}]$$ has been proven.
By the definition of subsets $2,$ $$(\mathrm{A} \cap \mathrm{B}) \subseteq \cap\{\mathrm{A}, \mathrm{B}\}$$ is
true.
Combining $1\;\&\;2,$ $\cap\{A,B\}=A\cap B$ has been proven.

Note: '$\bullet$' denotes the propositonal operator and
