Theorem resdmb

Description: Biconditional form of resdm , to replace it. (Contributed by BJ, 8-Nov-2026) TODO: this is dfrel5 : move to main section and update credits.

	Ref	Expression
Assertion	resdmb	$$⊢\mathrm{Rel}A\leftrightarrow {A\upharpoonright }_{\mathrm{dom}A}=A$$

Proof

Step	Hyp	Ref	Expression
1		ssid	$$⊢\mathrm{dom}A\subseteq \mathrm{dom}A$$
2	1	biantru	$$⊢\mathrm{Rel}A\leftrightarrow \left(\mathrm{Rel}A\wedge \mathrm{dom}A\subseteq \mathrm{dom}A\right)$$
3		relssresb	$$⊢\left(\mathrm{Rel}A\wedge \mathrm{dom}A\subseteq \mathrm{dom}A\right)\leftrightarrow {A\upharpoonright }_{\mathrm{dom}A}=A$$
4	2 3	bitri	$$⊢\mathrm{Rel}A\leftrightarrow {A\upharpoonright }_{\mathrm{dom}A}=A$$