Theorem relssdmrnb 38187

Description: Biconditional form of relssdmrn 6293, to replace it. (Contributed by BJ, 8-Nov-2026.) TODO: prove dfres3 5989 from it.

Assertion

Ref	Expression
relssdmrnb	⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (dom 𝐴 × ran 𝐴))

Proof of Theorem relssdmrnb

Step	Hyp	Ref	Expression
1		relssdmrn 6293	. 2 ⊢ (Rel 𝐴 → 𝐴 ⊆ (dom 𝐴 × ran 𝐴))
2		id 23	. . . 4 ⊢ (𝐴 ⊆ (dom 𝐴 × ran 𝐴) → 𝐴 ⊆ (dom 𝐴 × ran 𝐴))
3		xpss 5672	. . . 4 ⊢ (dom 𝐴 × ran 𝐴) ⊆ (V × V)
4	2, 3	sstrdi 3925	. . 3 ⊢ (𝐴 ⊆ (dom 𝐴 × ran 𝐴) → 𝐴 ⊆ (V × V))
5		df-rel 5663	. . 3 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
6	4, 5	sylibr 237	. 2 ⊢ (𝐴 ⊆ (dom 𝐴 × ran 𝐴) → Rel 𝐴)
7	1, 6	impbii 212	1 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (dom 𝐴 × ran 𝐴))

This proof depends on syntax axioms:   ↔ wb 209  Vcvv 3454   ⊆ wss 3900   × cxp 5654  dom cdm 5656  ran crn 5657  Rel wrel 5661

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5253  ax-pr 5396

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-xp 5662  df-rel 5663  df-cnv 5664  df-dm 5666  df-rn 5667

This theorem is used by:  relssresb  38188