Theorem bj-subst

Description: Proof of sbalex from core axioms, ax-10 (modal5), and bj-ax12 . (Contributed by BJ, 29-Dec-2020) (Proof modification is discouraged.)

	Ref	Expression
Assertion	bj-subst	|- ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )

Proof

Step	Hyp	Ref	Expression
1		bj-ax12	|- A. x ( x = y -> ( ph -> A. x ( x = y -> ph ) ) )
2		pm3.31	|- ( ( x = y -> ( ph -> A. x ( x = y -> ph ) ) ) -> ( ( x = y /\ ph ) -> A. x ( x = y -> ph ) ) )
3	2	aleximi	|- ( A. x ( x = y -> ( ph -> A. x ( x = y -> ph ) ) ) -> ( E. x ( x = y /\ ph ) -> E. x A. x ( x = y -> ph ) ) )
4	1 3	ax-mp	|- ( E. x ( x = y /\ ph ) -> E. x A. x ( x = y -> ph ) )
5		hbe1a	|- ( E. x A. x ( x = y -> ph ) -> A. x ( x = y -> ph ) )
6	4 5	syl	|- ( E. x ( x = y /\ ph ) -> A. x ( x = y -> ph ) )
7		equs4v	|- ( A. x ( x = y -> ph ) -> E. x ( x = y /\ ph ) )
8	6 7	impbii	|- ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )