Theorem bj-dfsbc

Description: Proof of df-sbc when taking bj-df-sb as definition. (Contributed by BJ, 19-Feb-2026) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	bj-dfsbc	$$⊢A\in \left\{x|\phi \right\}\leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$

Proof

Step	Hyp	Ref	Expression
1		df-clab	$$⊢y\in \left\{x|\phi \right\}\leftrightarrow \left[y/x\right]\phi$$
2		sb6	$$⊢\left[y/x\right]\phi \leftrightarrow \forall x\left(x=y\to \phi \right)$$
3	1 2	bitri	$$⊢y\in \left\{x|\phi \right\}\leftrightarrow \forall x\left(x=y\to \phi \right)$$
4	3	anbi2i	$$⊢\left(y=A\wedge y\in \left\{x|\phi \right\}\right)\leftrightarrow \left(y=A\wedge \forall x\left(x=y\to \phi \right)\right)$$
5	4	exbii	$$⊢\exists y\left(y=A\wedge y\in \left\{x|\phi \right\}\right)\leftrightarrow \exists y\left(y=A\wedge \forall x\left(x=y\to \phi \right)\right)$$
6		dfclel	$$⊢A\in \left\{x|\phi \right\}\leftrightarrow \exists y\left(y=A\wedge y\in \left\{x|\phi \right\}\right)$$
7		bj-df-sb	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=A\wedge \forall x\left(x=y\to \phi \right)\right)$$
8	5 6 7	3bitr4i	$$⊢A\in \left\{x|\phi \right\}\leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$