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Metamath Proof Explorer

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 ) )