Theorem. Show that if $\beta$ and $\beta'$ are two elements of a set $I$ and to each $\alpha \in I$ we assign a set $A_{\alpha}$, then $$ \{ x \in A_{\beta}: x \in A_{\alpha} \text{ for all } \alpha \in I \} = \{ x \in A_{\beta'}: x \in A_{\alpha} \text{ for all } \alpha \in I \}. $$


Since $\beta, \beta' \in I$ we have $(\forall \alpha \in I) \: x \in A_{\alpha} \rightarrow ((\exists \beta \in I) \: x \in A_{\beta}) \wedge ((\exists \beta' \in I) \: x \in A_{\beta'})$. This implies $$ ((\exists \beta \in I) \: y \in A_{\beta}) \wedge ((\forall \alpha \in I) \: y \in A_{\alpha} ) \leftrightarrow ((\exists \beta' \in I) \: y \in A_{\beta'}) \wedge ((\forall \alpha \in I) \: y \in A_{\alpha}). $$ According to the axiom of separation (3.5) this is equivalent to saying $$ y \in \{ x \in A_{\beta} : x \in A_{\alpha} \forall \alpha \in I \} \stackrel{(3.5)}{\longleftrightarrow} y \in A_{\beta} \wedge y \in \{ x : x \in A_{\alpha} \forall \alpha \in I \}. $$ $\Box$

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