Metamath Proof Explorer


Theorem oppfvallem

Description: Lemma for oppfval . (Contributed by Zhi Wang, 13-Nov-2025)

Ref Expression
Assertion oppfvallem ⊢ F C Func D G → Rel ⁡ G ∧ Rel ⁡ dom ⁡ G

Proof

Step Hyp Ref Expression
1 eqid ⊢ Base C = Base C
2 id ⊢ F C Func D G → F C Func D G
3 1 2 funcfn2 ⊢ F C Func D G → G Fn Base C × Base C
4 fnrel ⊢ G Fn Base C × Base C → Rel ⁡ G
5 3 4 syl ⊢ F C Func D G → Rel ⁡ G
6 relxp ⊢ Rel ⁡ Base C × Base C
7 3 fndmd ⊢ F C Func D G → dom ⁡ G = Base C × Base C
8 7 releqd ⊢ F C Func D G → Rel ⁡ dom ⁡ G ↔ Rel ⁡ Base C × Base C
9 6 8 mpbiri ⊢ F C Func D G → Rel ⁡ dom ⁡ G
10 5 9 jca ⊢ F C Func D G → Rel ⁡ G ∧ Rel ⁡ dom ⁡ G