Theorem oppfrcl3

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	|- ( ph -> G e. R )
	oppfrcl.2	|- Rel R
	oppfrcl.3	|- G = ( oppFunc ` F )
	oppfrcl2.4	|- ( ph -> F = <. A , B >. )
Assertion	oppfrcl3	|- ( ph -> ( Rel B /\ Rel dom B ) )

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	4	fveq2d	|- ( ph -> ( oppFunc ` F ) = ( oppFunc ` <. A , B >. ) )
6		df-ov	|- ( A oppFunc B ) = ( oppFunc ` <. A , B >. )
7	5 3 6	3eqtr4g	|- ( ph -> G = ( A oppFunc B ) )
8	1 2 3 4	oppfrcl2	|- ( ph -> ( A e. _V /\ B e. _V ) )
9		oppfvalg	|- ( ( A e. _V /\ B e. _V ) -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) )
10	8 9	syl	|- ( ph -> ( A oppFunc B ) = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) )
11	7 10	eqtrd	|- ( ph -> G = if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) )
12	1 2	oppfrcllem	|- ( ph -> G =/= (/) )
13	11 12	eqnetrrd	|- ( ph -> if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) =/= (/) )
14		iffalse	|- ( -. ( Rel B /\ Rel dom B ) -> if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) = (/) )
15	14	necon1ai	|- ( if ( ( Rel B /\ Rel dom B ) , <. A , tpos B >. , (/) ) =/= (/) -> ( Rel B /\ Rel dom B ) )
16	13 15	syl	|- ( ph -> ( Rel B /\ Rel dom B ) )