Metamath Proof Explorer


Theorem ceqsralv2

Description: Alternate elimination of a restricted universal quantifier, using implicit substitution. (Contributed by Scott Fenton, 7-Dec-2020)

Ref Expression
Hypothesis ceqsralv2.1
|- ( x = A -> ( ph <-> ps ) )
Assertion ceqsralv2
|- ( A. x e. B ( x = A -> ph ) <-> ( A e. B -> ps ) )

Proof

Step Hyp Ref Expression
1 ceqsralv2.1
 |-  ( x = A -> ( ph <-> ps ) )
2 1 notbid
 |-  ( x = A -> ( -. ph <-> -. ps ) )
3 2 ceqsrexv2
 |-  ( E. x e. B ( x = A /\ -. ph ) <-> ( A e. B /\ -. ps ) )
4 rexanali
 |-  ( E. x e. B ( x = A /\ -. ph ) <-> -. A. x e. B ( x = A -> ph ) )
5 annim
 |-  ( ( A e. B /\ -. ps ) <-> -. ( A e. B -> ps ) )
6 3 4 5 3bitr3i
 |-  ( -. A. x e. B ( x = A -> ph ) <-> -. ( A e. B -> ps ) )
7 6 con4bii
 |-  ( A. x e. B ( x = A -> ph ) <-> ( A e. B -> ps ) )