Theorem bj-sbcex

Description: Proof of sbcex 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-sbcex	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \to A\in \mathrm{V}$$

Proof

Step	Hyp	Ref	Expression
1		exsimpl	$$⊢\exists y\left(y=A\wedge \forall x\left(x=y\to \phi \right)\right)\to \exists yy=A$$
2		bj-df-sb	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=A\wedge \forall x\left(x=y\to \phi \right)\right)$$
3		isset	$$⊢A\in \mathrm{V}\leftrightarrow \exists yy=A$$
4	1 2 3	3imtr4i	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \to A\in \mathrm{V}$$