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 e. { x | ph } <-> [. A / x ]. ph )

Proof

Step	Hyp	Ref	Expression
1		df-clab	|- ( y e. { x | ph } <-> [ y / x ] ph )
2		sb6	|- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
3	1 2	bitri	|- ( y e. { x | ph } <-> A. x ( x = y -> ph ) )
4	3	anbi2i	|- ( ( y = A /\ y e. { x | ph } ) <-> ( y = A /\ A. x ( x = y -> ph ) ) )
5	4	exbii	|- ( E. y ( y = A /\ y e. { x | ph } ) <-> E. y ( y = A /\ A. x ( x = y -> ph ) ) )
6		dfclel	|- ( A e. { x | ph } <-> E. y ( y = A /\ y e. { x | ph } ) )
7		bj-df-sb	|- ( [. A / x ]. ph <-> E. y ( y = A /\ A. x ( x = y -> ph ) ) )
8	5 6 7	3bitr4i	|- ( A e. { x | ph } <-> [. A / x ]. ph )