Metamath Proof Explorer


Theorem equsexvw

Description: Version of equsexv with a disjoint variable condition, and of equsex with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsalvw . (Contributed by BJ, 31-May-2019) (Proof shortened by Wolf Lammen, 23-Oct-2023)

Ref Expression
Hypothesis equsalvw.1
|- ( x = y -> ( ph <-> ps ) )
Assertion equsexvw
|- ( E. x ( x = y /\ ph ) <-> ps )

Proof

Step Hyp Ref Expression
1 equsalvw.1
 |-  ( x = y -> ( ph <-> ps ) )
2 alinexa
 |-  ( A. x ( x = y -> -. ph ) <-> -. E. x ( x = y /\ ph ) )
3 1 notbid
 |-  ( x = y -> ( -. ph <-> -. ps ) )
4 3 equsalvw
 |-  ( A. x ( x = y -> -. ph ) <-> -. ps )
5 2 4 bitr3i
 |-  ( -. E. x ( x = y /\ ph ) <-> -. ps )
6 5 con4bii
 |-  ( E. x ( x = y /\ ph ) <-> ps )