Metamath Proof Explorer


Theorem oppfval2

Description: Value of the opposite functor. (Contributed by Zhi Wang, 13-Nov-2025)

Ref Expression
Assertion oppfval2 ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝐹 ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ )

Proof

Step Hyp Ref Expression
1 relfunc Rel ( 𝐶 Func 𝐷 )
2 1st2nd ( ( Rel ( 𝐶 Func 𝐷 ) ∧ 𝐹 ∈ ( 𝐶 Func 𝐷 ) ) → 𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
3 1 2 mpan ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → 𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
4 3 fveq2d ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝐹 ) = ( oppFunc ‘ ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ) )
5 df-ov ( ( 1st𝐹 ) oppFunc ( 2nd𝐹 ) ) = ( oppFunc ‘ ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
6 4 5 eqtr4di ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝐹 ) = ( ( 1st𝐹 ) oppFunc ( 2nd𝐹 ) ) )
7 1st2ndbr ( ( Rel ( 𝐶 Func 𝐷 ) ∧ 𝐹 ∈ ( 𝐶 Func 𝐷 ) ) → ( 1st𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd𝐹 ) )
8 1 7 mpan ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( 1st𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd𝐹 ) )
9 oppfval ( ( 1st𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd𝐹 ) → ( ( 1st𝐹 ) oppFunc ( 2nd𝐹 ) ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ )
10 8 9 syl ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( ( 1st𝐹 ) oppFunc ( 2nd𝐹 ) ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ )
11 6 10 eqtrd ( 𝐹 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝐹 ) = ⟨ ( 1st𝐹 ) , tpos ( 2nd𝐹 ) ⟩ )