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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 
|- ( ph -> G e. R )
oppfrcl.2 
|- Rel R
oppfrcl.3 
|- G = ( oppFunc ` F )
oppfrcl2.4 
|- ( ph -> F = <. A , B >. )
Assertion oppfrcl2 
|- ( ph -> ( A e. _V /\ B e. _V ) )

Proof

Step Hyp Ref Expression
1 oppfrcl.1 
 |-  ( ph -> G e. R )
2 oppfrcl.2 
 |-  Rel R
3 oppfrcl.3 
 |-  G = ( oppFunc ` F )
4 oppfrcl2.4 
 |-  ( ph -> F = <. A , B >. )
5 1 2 3 oppfrcl 
 |-  ( ph -> F e. ( _V X. _V ) )
6 4 5 eqeltrrd 
 |-  ( ph -> <. A , B >. e. ( _V X. _V ) )
7 0nelxp 
 |-  -. (/) e. ( _V X. _V )
8 nelne2 
 |-  ( ( <. A , B >. e. ( _V X. _V ) /\ -. (/) e. ( _V X. _V ) ) -> <. A , B >. =/= (/) )
9 6 7 8 sylancl 
 |-  ( ph -> <. A , B >. =/= (/) )
10 opprc 
 |-  ( -. ( A e. _V /\ B e. _V ) -> <. A , B >. = (/) )
11 10 necon1ai 
 |-  ( <. A , B >. =/= (/) -> ( A e. _V /\ B e. _V ) )
12 9 11 syl 
 |-  ( ph -> ( A e. _V /\ B e. _V ) )