Theorem eliminable-veqab

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

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

Proof

Step	Hyp	Ref	Expression
1		dfcleq	$$⊢x=\left\{y|\phi \right\}\leftrightarrow \forall z\left(z\in x\leftrightarrow z\in \left\{y|\phi \right\}\right)$$
2		eliminable-velab	$$⊢z\in \left\{y|\phi \right\}\leftrightarrow \left[z/y\right]\phi$$
3	2	bibi2i	$$⊢\left(z\in x\leftrightarrow z\in \left\{y|\phi \right\}\right)\leftrightarrow \left(z\in x\leftrightarrow \left[z/y\right]\phi \right)$$
4	3	albii	$$⊢\forall z\left(z\in x\leftrightarrow z\in \left\{y|\phi \right\}\right)\leftrightarrow \forall z\left(z\in x\leftrightarrow \left[z/y\right]\phi \right)$$
5	1 4	bitri	$$⊢x=\left\{y|\phi \right\}\leftrightarrow \forall z\left(z\in x\leftrightarrow \left[z/y\right]\phi \right)$$