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Metamath Proof Explorer

Theorem hbim

Description: If x is not free in ph and ps , it is not free in ( ph -> ps ) . (Contributed by NM, 24-Jan-1993) (Proof shortened by Mel L. O'Cat, 3-Mar-2008) (Proof shortened by Wolf Lammen, 1-Jan-2018)

Ref Expression
Hypotheses hbim.1 
|- ( ph -> A. x ph )
hbim.2 
|- ( ps -> A. x ps )
Assertion hbim 
|- ( ( ph -> ps ) -> A. x ( ph -> ps ) )

Proof

Step Hyp Ref Expression
1 hbim.1 
 |-  ( ph -> A. x ph )
2 hbim.2 
 |-  ( ps -> A. x ps )
3 2 a1i 
 |-  ( ph -> ( ps -> A. x ps ) )
4 1 3 hbim1 
 |-  ( ( ph -> ps ) -> A. x ( ph -> ps ) )