Metamath Proof Explorer


Theorem opf2fval

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

Ref Expression
Hypotheses opf2fval.f ( 𝜑𝐹 = ( 𝑥𝐴 , 𝑦𝐵 ↦ ( I ↾ ( 𝑦 𝑁 𝑥 ) ) ) )
opf2fval.x ( 𝜑𝑋𝐴 )
opf2fval.y ( 𝜑𝑌𝐵 )
Assertion opf2fval ( 𝜑 → ( 𝑋 𝐹 𝑌 ) = ( I ↾ ( 𝑌 𝑁 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 opf2fval.f ( 𝜑𝐹 = ( 𝑥𝐴 , 𝑦𝐵 ↦ ( I ↾ ( 𝑦 𝑁 𝑥 ) ) ) )
2 opf2fval.x ( 𝜑𝑋𝐴 )
3 opf2fval.y ( 𝜑𝑌𝐵 )
4 simprr ( ( 𝜑 ∧ ( 𝑥 = 𝑋𝑦 = 𝑌 ) ) → 𝑦 = 𝑌 )
5 simprl ( ( 𝜑 ∧ ( 𝑥 = 𝑋𝑦 = 𝑌 ) ) → 𝑥 = 𝑋 )
6 4 5 oveq12d ( ( 𝜑 ∧ ( 𝑥 = 𝑋𝑦 = 𝑌 ) ) → ( 𝑦 𝑁 𝑥 ) = ( 𝑌 𝑁 𝑋 ) )
7 6 reseq2d ( ( 𝜑 ∧ ( 𝑥 = 𝑋𝑦 = 𝑌 ) ) → ( I ↾ ( 𝑦 𝑁 𝑥 ) ) = ( I ↾ ( 𝑌 𝑁 𝑋 ) ) )
8 ovexd ( 𝜑 → ( 𝑌 𝑁 𝑋 ) ∈ V )
9 resiexg ( ( 𝑌 𝑁 𝑋 ) ∈ V → ( I ↾ ( 𝑌 𝑁 𝑋 ) ) ∈ V )
10 8 9 syl ( 𝜑 → ( I ↾ ( 𝑌 𝑁 𝑋 ) ) ∈ V )
11 1 7 2 3 10 ovmpod ( 𝜑 → ( 𝑋 𝐹 𝑌 ) = ( I ↾ ( 𝑌 𝑁 𝑋 ) ) )