Metamath Proof Explorer


Theorem spei

Description: Inference from existential specialization, using implicit substitution. Remove a distinct variable constraint. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker speiv if possible. (Contributed by NM, 19-Aug-1993) (Proof shortened by Wolf Lammen, 12-May-2018) (New usage is discouraged.)

Ref Expression
Hypotheses spei.1
|- ( x = y -> ( ph <-> ps ) )
spei.2
|- ps
Assertion spei
|- E. x ph

Proof

Step Hyp Ref Expression
1 spei.1
 |-  ( x = y -> ( ph <-> ps ) )
2 spei.2
 |-  ps
3 ax6e
 |-  E. x x = y
4 2 1 mpbiri
 |-  ( x = y -> ph )
5 3 4 eximii
 |-  E. x ph