Metamath Proof Explorer


Theorem oppfval

Description: Value of the opposite functor. (Contributed by Zhi Wang, 4-Nov-2025)

Ref Expression
Assertion oppfval ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 → ( 𝐹 oppFunc 𝐺 ) = ⟨ 𝐹 , tpos 𝐺 ⟩ )

Proof

Step Hyp Ref Expression
1 relfunc Rel ( 𝐶 Func 𝐷 )
2 1 brrelex12i ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 → ( 𝐹 ∈ V ∧ 𝐺 ∈ V ) )
3 simpl ( ( 𝑓 = 𝐹𝑔 = 𝐺 ) → 𝑓 = 𝐹 )
4 simpr ( ( 𝑓 = 𝐹𝑔 = 𝐺 ) → 𝑔 = 𝐺 )
5 4 tposeqd ( ( 𝑓 = 𝐹𝑔 = 𝐺 ) → tpos 𝑔 = tpos 𝐺 )
6 3 5 opeq12d ( ( 𝑓 = 𝐹𝑔 = 𝐺 ) → ⟨ 𝑓 , tpos 𝑔 ⟩ = ⟨ 𝐹 , tpos 𝐺 ⟩ )
7 df-oppf oppFunc = ( 𝑓 ∈ V , 𝑔 ∈ V ↦ ⟨ 𝑓 , tpos 𝑔 ⟩ )
8 opex 𝐹 , tpos 𝐺 ⟩ ∈ V
9 6 7 8 ovmpoa ( ( 𝐹 ∈ V ∧ 𝐺 ∈ V ) → ( 𝐹 oppFunc 𝐺 ) = ⟨ 𝐹 , tpos 𝐺 ⟩ )
10 2 9 syl ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 → ( 𝐹 oppFunc 𝐺 ) = ⟨ 𝐹 , tpos 𝐺 ⟩ )