Metamath Proof Explorer


Theorem relssdmrnb

Description: Biconditional form of relssdmrn , to replace it. (Contributed by BJ, 8-Nov-2026) TODO: prove dfres3 from it.

Ref Expression
Assertion relssdmrnb ⊢ Rel ⁡ A ↔ A ⊆ dom ⁡ A × ran ⁡ A

Proof

Step Hyp Ref Expression
1 relssdmrn ⊢ Rel ⁡ A → A ⊆ dom ⁡ A × ran ⁡ A
2 id ⊢ A ⊆ dom ⁡ A × ran ⁡ A → A ⊆ dom ⁡ A × ran ⁡ A
3 xpss ⊢ dom ⁡ A × ran ⁡ A ⊆ V × V
4 2 3 sstrdi ⊢ A ⊆ dom ⁡ A × ran ⁡ A → A ⊆ V × V
5 df-rel ⊢ Rel ⁡ A ↔ A ⊆ V × V
6 4 5 sylibr ⊢ A ⊆ dom ⁡ A × ran ⁡ A → Rel ⁡ A
7 1 6 impbii ⊢ Rel ⁡ A ↔ A ⊆ dom ⁡ A × ran ⁡ A