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	|- ( [. A / x ]. ph -> A e. _V )

Proof

Step	Hyp	Ref	Expression
1		exsimpl	|- ( E. y ( y = A /\ A. x ( x = y -> ph ) ) -> E. y y = A )
2		bj-df-sb	|- ( [. A / x ]. ph <-> E. y ( y = A /\ A. x ( x = y -> ph ) ) )
3		isset	|- ( A e. _V <-> E. y y = A )
4	1 2 3	3imtr4i	|- ( [. A / x ]. ph -> A e. _V )