Metamath Proof Explorer


Theorem oppf2

Description: Value of the morphism part of the opposite functor. (Contributed by Zhi Wang, 19-Nov-2025)

Ref Expression
Hypothesis oppf1.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
Assertion oppf2 ( 𝜑 → ( 𝑀 ( 2nd ‘ ( oppFunc ‘ 𝐹 ) ) 𝑁 ) = ( 𝑁 ( 2nd𝐹 ) 𝑀 ) )

Proof

Step Hyp Ref Expression
1 oppf1.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
2 oppfval2 ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝐹 ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ )
3 fvex ( 1st𝐹 ) ∈ V
4 fvex ( 2nd𝐹 ) ∈ V
5 4 tposex tpos ( 2nd𝐹 ) ∈ V
6 3 5 op2ndd ( ( oppFunc ‘ 𝐹 ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ → ( 2nd ‘ ( oppFunc ‘ 𝐹 ) ) = tpos ( 2nd𝐹 ) )
7 1 2 6 3syl ( 𝜑 → ( 2nd ‘ ( oppFunc ‘ 𝐹 ) ) = tpos ( 2nd𝐹 ) )
8 7 oveqd ( 𝜑 → ( 𝑀 ( 2nd ‘ ( oppFunc ‘ 𝐹 ) ) 𝑁 ) = ( 𝑀 tpos ( 2nd𝐹 ) 𝑁 ) )
9 ovtpos ( 𝑀 tpos ( 2nd𝐹 ) 𝑁 ) = ( 𝑁 ( 2nd𝐹 ) 𝑀 )
10 8 9 eqtrdi ( 𝜑 → ( 𝑀 ( 2nd ‘ ( oppFunc ‘ 𝐹 ) ) 𝑁 ) = ( 𝑁 ( 2nd𝐹 ) 𝑀 ) )