Metamath Proof Explorer


Theorem oppfrcl2

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

Ref Expression
Hypotheses oppfrcl.1 ( 𝜑𝐺𝑅 )
oppfrcl.2 Rel 𝑅
oppfrcl.3 𝐺 = ( oppFunc ‘ 𝐹 )
oppfrcl2.4 ( 𝜑𝐹 = ⟨ 𝐴 , 𝐵 ⟩ )
Assertion oppfrcl2 ( 𝜑 → ( 𝐴 ∈ V ∧ 𝐵 ∈ V ) )

Proof

Step Hyp Ref Expression
1 oppfrcl.1 ( 𝜑𝐺𝑅 )
2 oppfrcl.2 Rel 𝑅
3 oppfrcl.3 𝐺 = ( oppFunc ‘ 𝐹 )
4 oppfrcl2.4 ( 𝜑𝐹 = ⟨ 𝐴 , 𝐵 ⟩ )
5 1 2 3 oppfrcl ( 𝜑𝐹 ∈ ( V × V ) )
6 4 5 eqeltrrd ( 𝜑 → ⟨ 𝐴 , 𝐵 ⟩ ∈ ( V × V ) )
7 0nelxp ¬ ∅ ∈ ( V × V )
8 nelne2 ( ( ⟨ 𝐴 , 𝐵 ⟩ ∈ ( V × V ) ∧ ¬ ∅ ∈ ( V × V ) ) → ⟨ 𝐴 , 𝐵 ⟩ ≠ ∅ )
9 6 7 8 sylancl ( 𝜑 → ⟨ 𝐴 , 𝐵 ⟩ ≠ ∅ )
10 opprc ( ¬ ( 𝐴 ∈ V ∧ 𝐵 ∈ V ) → ⟨ 𝐴 , 𝐵 ⟩ = ∅ )
11 10 necon1ai ( ⟨ 𝐴 , 𝐵 ⟩ ≠ ∅ → ( 𝐴 ∈ V ∧ 𝐵 ∈ V ) )
12 9 11 syl ( 𝜑 → ( 𝐴 ∈ V ∧ 𝐵 ∈ V ) )