Metamath Proof Explorer


Theorem oppchomfpropd

Description: If two categories have the same hom-sets, so do their opposites. (Contributed by Mario Carneiro, 26-Jan-2017)

Ref Expression
Hypothesis oppchomfpropd.1 ( 𝜑 → ( Homf𝐶 ) = ( Homf𝐷 ) )
Assertion oppchomfpropd ( 𝜑 → ( Homf ‘ ( oppCat ‘ 𝐶 ) ) = ( Homf ‘ ( oppCat ‘ 𝐷 ) ) )

Proof

Step Hyp Ref Expression
1 oppchomfpropd.1 ( 𝜑 → ( Homf𝐶 ) = ( Homf𝐷 ) )
2 1 tposeqd ( 𝜑 → tpos ( Homf𝐶 ) = tpos ( Homf𝐷 ) )
3 eqid ( oppCat ‘ 𝐶 ) = ( oppCat ‘ 𝐶 )
4 eqid ( Homf𝐶 ) = ( Homf𝐶 )
5 3 4 oppchomf tpos ( Homf𝐶 ) = ( Homf ‘ ( oppCat ‘ 𝐶 ) )
6 eqid ( oppCat ‘ 𝐷 ) = ( oppCat ‘ 𝐷 )
7 eqid ( Homf𝐷 ) = ( Homf𝐷 )
8 6 7 oppchomf tpos ( Homf𝐷 ) = ( Homf ‘ ( oppCat ‘ 𝐷 ) )
9 2 5 8 3eqtr3g ( 𝜑 → ( Homf ‘ ( oppCat ‘ 𝐶 ) ) = ( Homf ‘ ( oppCat ‘ 𝐷 ) ) )