Metamath Proof Explorer


Theorem oppfrcl

Description: If an opposite functor of a class is a functor, then the original class must be an ordered pair. (Contributed by Zhi Wang, 14-Nov-2025)

Ref Expression
Hypotheses oppfrcl.1 ⊢ ( 𝜑 → 𝐺 ∈ 𝑅 )
oppfrcl.2 ⊢ Rel 𝑅
oppfrcl.3 ⊢ 𝐺 = ( oppFunc ‘ 𝐹 )
Assertion oppfrcl ( 𝜑 → 𝐹 ∈ ( V × V ) )

Proof

Step Hyp Ref Expression
1 oppfrcl.1 ⊢ ( 𝜑 → 𝐺 ∈ 𝑅 )
2 oppfrcl.2 ⊢ Rel 𝑅
3 oppfrcl.3 ⊢ 𝐺 = ( oppFunc ‘ 𝐹 )
4 1 2 oppfrcllem ⊢ ( 𝜑 → 𝐺 ≠ ∅ )
5 ndmfv ⊢ ( ¬ 𝐹 ∈ dom oppFunc → ( oppFunc ‘ 𝐹 ) = ∅ )
6 3 5 eqtrid ⊢ ( ¬ 𝐹 ∈ dom oppFunc → 𝐺 = ∅ )
7 6 necon1ai ⊢ ( 𝐺 ≠ ∅ → 𝐹 ∈ dom oppFunc )
8 4 7 syl ⊢ ( 𝜑 → 𝐹 ∈ dom oppFunc )
9 oppffn ⊢ oppFunc Fn ( V × V )
10 9 fndmi ⊢ dom oppFunc = ( V × V )
11 8 10 eleqtrdi ⊢ ( 𝜑 → 𝐹 ∈ ( V × V ) )