Metamath Proof Explorer

Theorem ancomd

Description: Commutation of conjuncts in consequent. (Contributed by Jeff Hankins, 14-Aug-2009)

Ref Expression
Hypothesis ancomd.1 
|- ( ph -> ( ps /\ ch ) )
Assertion ancomd 
|- ( ph -> ( ch /\ ps ) )

Proof

Step Hyp Ref Expression
1 ancomd.1 
 |-  ( ph -> ( ps /\ ch ) )
2 ancom 
 |-  ( ( ps /\ ch ) <-> ( ch /\ ps ) )
3 1 2 sylib 
 |-  ( ph -> ( ch /\ ps ) )