Metamath Proof Explorer

Theorem bj-spst

Description: Closed form of sps . Once in main part, prove sps and spsd from it. (Contributed by BJ, 20-Oct-2019)

Ref Expression
Assertion bj-spst 
|- ( ( ph -> ps ) -> ( A. x ph -> ps ) )

Proof

Step Hyp Ref Expression
1 sp 
 |-  ( A. x ph -> ph )
2 1 imim1i 
 |-  ( ( ph -> ps ) -> ( A. x ph -> ps ) )