Metamath Proof Explorer


Theorem spv

Description: Specialization, using implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker spvv if possible. (Contributed by NM, 30-Aug-1993) (New usage is discouraged.)

Ref Expression
Hypothesis spv.1
|- ( x = y -> ( ph <-> ps ) )
Assertion spv
|- ( A. x ph -> ps )

Proof

Step Hyp Ref Expression
1 spv.1
 |-  ( x = y -> ( ph <-> ps ) )
2 1 biimpd
 |-  ( x = y -> ( ph -> ps ) )
3 2 spimv
 |-  ( A. x ph -> ps )