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Description: If an opposite functor of a class is a functor, then the second component of the original class must be a relation whose domain is a relation as well. (Contributed by Zhi Wang, 14-Nov-2025)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | oppfrcl.1 | ||
| oppfrcl.2 | |||
| oppfrcl.3 | No typesetting found for |- G = ( oppFunc ` F ) with typecode |- | ||
| oppfrcl2.4 | |||
| Assertion | oppfrcl3 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | oppfrcl.1 | ||
| 2 | oppfrcl.2 | ||
| 3 | oppfrcl.3 | Could not format G = ( oppFunc ` F ) : No typesetting found for |- G = ( oppFunc ` F ) with typecode |- | |
| 4 | oppfrcl2.4 | ||
| 5 | 4 | fveq2d | Could not format ( ph -> ( oppFunc ` F ) = ( oppFunc ` <. A , B >. ) ) : No typesetting found for |- ( ph -> ( oppFunc ` F ) = ( oppFunc ` <. A , B >. ) ) with typecode |- |
| 6 | df-ov | Could not format ( A oppFunc B ) = ( oppFunc ` <. A , B >. ) : No typesetting found for |- ( A oppFunc B ) = ( oppFunc ` <. A , B >. ) with typecode |- | |
| 7 | 5 3 6 | 3eqtr4g | Could not format ( ph -> G = ( A oppFunc B ) ) : No typesetting found for |- ( ph -> G = ( A oppFunc B ) ) with typecode |- |
| 8 | 1 2 3 4 | oppfrcl2 | |
| 9 | oppfvalg | Could not format ( ( A e. _V /\ B e. _V ) -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) ) : No typesetting found for |- ( ( A e. _V /\ B e. _V ) -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) ) with typecode |- | |
| 10 | 8 9 | syl | Could not format ( ph -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) ) : No typesetting found for |- ( ph -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) ) with typecode |- |
| 11 | 7 10 | eqtrd | |
| 12 | 1 2 | oppfrcllem | |
| 13 | 11 12 | eqnetrrd | |
| 14 | iffalse | ||
| 15 | 14 | necon1ai | |
| 16 | 13 15 | syl |