Metamath Proof Explorer


Theorem consym1

Description: A symmetry with /\ .

See negsym1 for more information. (Contributed by Anthony Hart, 4-Sep-2011)

Ref Expression
Assertion consym1
|- ( ( ps /\ ( ps /\ F. ) ) -> ( ps /\ ph ) )

Proof

Step Hyp Ref Expression
1 falim
 |-  ( F. -> ( ( ps /\ ( ps /\ F. ) ) -> ( ps /\ ph ) ) )
2 1 ad2antll
 |-  ( ( ps /\ ( ps /\ F. ) ) -> ( ( ps /\ ( ps /\ F. ) ) -> ( ps /\ ph ) ) )
3 2 pm2.43i
 |-  ( ( ps /\ ( ps /\ F. ) ) -> ( ps /\ ph ) )