Theorem refreleq 34585

Description: Equality theorem for reflexive relation. (Contributed by Peter Mazsa, 15-Apr-2019.) (Revised by Peter Mazsa, 23-Sep-2021.)

Assertion

Ref	Expression
refreleq	⊢ (𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆))

Proof of Theorem refreleq

Step	Hyp	Ref	Expression
1		dmeq 5471	. . . . . 6 ⊢ (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
2		rneq 5498	. . . . . 6 ⊢ (𝑅 = 𝑆 → ran 𝑅 = ran 𝑆)
3	1, 2	xpeq12d 5289	. . . . 5 ⊢ (𝑅 = 𝑆 → (dom 𝑅 × ran 𝑅) = (dom 𝑆 × ran 𝑆))
4	3	ineq2d 3949	. . . 4 ⊢ (𝑅 = 𝑆 → ( I ∩ (dom 𝑅 × ran 𝑅)) = ( I ∩ (dom 𝑆 × ran 𝑆)))
5		id 22	. . . 4 ⊢ (𝑅 = 𝑆 → 𝑅 = 𝑆)
6	4, 5	sseq12d 3767	. . 3 ⊢ (𝑅 = 𝑆 → (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ↔ ( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆))
7		releq 5350	. . 3 ⊢ (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
8	6, 7	anbi12d 749	. 2 ⊢ (𝑅 = 𝑆 → ((( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅) ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆)))
9		dfrefrel2 34580	. 2 ⊢ ( RefRel 𝑅 ↔ (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅))
10		dfrefrel2 34580	. 2 ⊢ ( RefRel 𝑆 ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆))
11	8, 9, 10	3bitr4g 303	1 ⊢ (𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 383   = wceq 1624   ∩ cin 3706   ⊆ wss 3707   I cid 5165   × cxp 5256  dom cdm 5258  ran crn 5259  Rel wrel 5263   RefRel wrefrel 34294

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1863  ax-4 1878  ax-5 1980  ax-6 2046  ax-7 2082  ax-9 2140  ax-10 2160  ax-11 2175  ax-12 2188  ax-13 2383  ax-ext 2732  ax-sep 4925  ax-nul 4933  ax-pr 5047

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1627  df-ex 1846  df-nf 1851  df-sb 2039  df-eu 2603  df-mo 2604  df-clab 2739  df-cleq 2745  df-clel 2748  df-nfc 2883  df-ral 3047  df-rex 3048  df-rab 3051  df-v 3334  df-dif 3710  df-un 3712  df-in 3714  df-ss 3721  df-nul 4051  df-if 4223  df-sn 4314  df-pr 4316  df-op 4320  df-br 4797  df-opab 4857  df-xp 5264  df-rel 5265  df-cnv 5266  df-dm 5268  df-rn 5269  df-res 5270  df-refrel 34577

This theorem is referenced by: (None)