Metamath Proof Explorer


Theorem spcdv

Description: Rule of specialization, using implicit substitution. Analogous to rspcdv . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypotheses spcimdv.1
|- ( ph -> A e. B )
spcdv.2
|- ( ( ph /\ x = A ) -> ( ps <-> ch ) )
Assertion spcdv
|- ( ph -> ( A. x ps -> ch ) )

Proof

Step Hyp Ref Expression
1 spcimdv.1
 |-  ( ph -> A e. B )
2 spcdv.2
 |-  ( ( ph /\ x = A ) -> ( ps <-> ch ) )
3 2 biimpd
 |-  ( ( ph /\ x = A ) -> ( ps -> ch ) )
4 1 3 spcimdv
 |-  ( ph -> ( A. x ps -> ch ) )