Theorem eliminable-abeqab

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): abstraction equals abstraction. (Contributed by BJ, 30-Apr-2024) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	eliminable-abeqab	$$⊢\left\{x|\phi \right\}=\left\{y|\psi \right\}\leftrightarrow \forall z\left(\left[z/x\right]\phi \leftrightarrow \left[z/y\right]\psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		dfcleq	$$⊢\left\{x|\phi \right\}=\left\{y|\psi \right\}\leftrightarrow \forall z\left(z\in \left\{x|\phi \right\}\leftrightarrow z\in \left\{y|\psi \right\}\right)$$
2		eliminable-velab	$$⊢z\in \left\{x|\phi \right\}\leftrightarrow \left[z/x\right]\phi$$
3		eliminable-velab	$$⊢z\in \left\{y|\psi \right\}\leftrightarrow \left[z/y\right]\psi$$
4	2 3	bibi12i	$$⊢\left(z\in \left\{x|\phi \right\}\leftrightarrow z\in \left\{y|\psi \right\}\right)\leftrightarrow \left(\left[z/x\right]\phi \leftrightarrow \left[z/y\right]\psi \right)$$
5	4	albii	$$⊢\forall z\left(z\in \left\{x|\phi \right\}\leftrightarrow z\in \left\{y|\psi \right\}\right)\leftrightarrow \forall z\left(\left[z/x\right]\phi \leftrightarrow \left[z/y\right]\psi \right)$$
6	1 5	bitri	$$⊢\left\{x|\phi \right\}=\left\{y|\psi \right\}\leftrightarrow \forall z\left(\left[z/x\right]\phi \leftrightarrow \left[z/y\right]\psi \right)$$