Theorem refreleq 36617

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 5809	. . . . . 6 ⊢ (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
2		rneq 5842	. . . . . 6 ⊢ (𝑅 = 𝑆 → ran 𝑅 = ran 𝑆)
3	1, 2	xpeq12d 5619	. . . . 5 ⊢ (𝑅 = 𝑆 → (dom 𝑅 × ran 𝑅) = (dom 𝑆 × ran 𝑆))
4	3	ineq2d 4151	. . . 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 3958	. . 3 ⊢ (𝑅 = 𝑆 → (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ↔ ( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆))
7		releq 5685	. . 3 ⊢ (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
8	6, 7	anbi12d 630	. 2 ⊢ (𝑅 = 𝑆 → ((( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅) ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆)))
9		dfrefrel2 36612	. 2 ⊢ ( RefRel 𝑅 ↔ (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅))
10		dfrefrel2 36612	. 2 ⊢ ( RefRel 𝑆 ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆))
11	8, 9, 10	3bitr4g 313	1 ⊢ (𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1541   ∩ cin 3890   ⊆ wss 3891   I cid 5487   × cxp 5586  dom cdm 5588  ran crn 5589  Rel wrel 5593   RefRel wrefrel 36318

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-10 2140  ax-12 2174  ax-ext 2710  ax-sep 5226  ax-nul 5233  ax-pr 5355

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1544  df-fal 1554  df-ex 1786  df-nf 1790  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-ral 3070  df-rex 3071  df-rab 3074  df-v 3432  df-dif 3894  df-un 3896  df-in 3898  df-ss 3908  df-nul 4262  df-if 4465  df-sn 4567  df-pr 4569  df-op 4573  df-br 5079  df-opab 5141  df-xp 5594  df-rel 5595  df-cnv 5596  df-dm 5598  df-rn 5599  df-res 5600  df-refrel 36609

This theorem is referenced by:  eqvreleq  36694