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 ( 𝜑𝐺𝑅 )
oppfrcl.2 Rel 𝑅
oppfrcl.3 𝐺 = ( oppFunc ‘ 𝐹 )
Assertion oppfrcl ( 𝜑𝐹 ∈ ( V × V ) )

Proof

Step Hyp Ref Expression
1 oppfrcl.1 ( 𝜑𝐺𝑅 )
2 oppfrcl.2 Rel 𝑅
3 oppfrcl.3 𝐺 = ( oppFunc ‘ 𝐹 )
4 1 2 oppfrcllem ( 𝜑𝐺 ≠ ∅ )
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 ( 𝜑𝐹 ∈ dom oppFunc )
9 oppffn oppFunc Fn ( V × V )
10 9 fndmi dom oppFunc = ( V × V )
11 8 10 eleqtrdi ( 𝜑𝐹 ∈ ( V × V ) )