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Metamath Proof Explorer

Theorem spvv

Description: Specialization, using implicit substitution. Version of spv with a disjoint variable condition, which does not require ax-7 , ax-12 , ax-13 . (Contributed by NM, 30-Aug-1993) (Revised by BJ, 31-May-2019)

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

Proof

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