Metamath Proof Explorer


Theorem opf11

Description: The object part of the op functor on functor categories. Lemma for fucoppc . (Contributed by Zhi Wang, 18-Nov-2025)

Ref Expression
Hypotheses opf11.f ( 𝜑𝐹 = ( oppFunc ↾ ( 𝐶 Func 𝐷 ) ) )
opf11.x ( 𝜑𝑋 ∈ ( 𝐶 Func 𝐷 ) )
Assertion opf11 ( 𝜑 → ( 1st ‘ ( 𝐹𝑋 ) ) = ( 1st𝑋 ) )

Proof

Step Hyp Ref Expression
1 opf11.f ( 𝜑𝐹 = ( oppFunc ↾ ( 𝐶 Func 𝐷 ) ) )
2 opf11.x ( 𝜑𝑋 ∈ ( 𝐶 Func 𝐷 ) )
3 1 fveq1d ( 𝜑 → ( 𝐹𝑋 ) = ( ( oppFunc ↾ ( 𝐶 Func 𝐷 ) ) ‘ 𝑋 ) )
4 2 fvresd ( 𝜑 → ( ( oppFunc ↾ ( 𝐶 Func 𝐷 ) ) ‘ 𝑋 ) = ( oppFunc ‘ 𝑋 ) )
5 oppfval2 ( 𝑋 ∈ ( 𝐶 Func 𝐷 ) → ( oppFunc ‘ 𝑋 ) = ⟨ ( 1st𝑋 ) , tpos ( 2nd𝑋 ) ⟩ )
6 2 5 syl ( 𝜑 → ( oppFunc ‘ 𝑋 ) = ⟨ ( 1st𝑋 ) , tpos ( 2nd𝑋 ) ⟩ )
7 3 4 6 3eqtrd ( 𝜑 → ( 𝐹𝑋 ) = ⟨ ( 1st𝑋 ) , tpos ( 2nd𝑋 ) ⟩ )
8 fvex ( 1st𝑋 ) ∈ V
9 fvex ( 2nd𝑋 ) ∈ V
10 9 tposex tpos ( 2nd𝑋 ) ∈ V
11 8 10 op1std ( ( 𝐹𝑋 ) = ⟨ ( 1st𝑋 ) , tpos ( 2nd𝑋 ) ⟩ → ( 1st ‘ ( 𝐹𝑋 ) ) = ( 1st𝑋 ) )
12 7 11 syl ( 𝜑 → ( 1st ‘ ( 𝐹𝑋 ) ) = ( 1st𝑋 ) )