Theorem refreleq 39283

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 5895	. . . . . 6 ⊢ (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
2		rneq 5928	. . . . . 6 ⊢ (𝑅 = 𝑆 → ran 𝑅 = ran 𝑆)
3	1, 2	xpeq12d 5694	. . . . 5 ⊢ (𝑅 = 𝑆 → (dom 𝑅 × ran 𝑅) = (dom 𝑆 × ran 𝑆))
4	3	ineq2d 4173	. . . 4 ⊢ (𝑅 = 𝑆 → ( I ∩ (dom 𝑅 × ran 𝑅)) = ( I ∩ (dom 𝑆 × ran 𝑆)))
5		id 23	. . . 4 ⊢ (𝑅 = 𝑆 → 𝑅 = 𝑆)
6	4, 5	sseq12d 3971	. . 3 ⊢ (𝑅 = 𝑆 → (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ↔ ( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆))
7		releq 5765	. . 3 ⊢ (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
8	6, 7	anbi12d 644	. 2 ⊢ (𝑅 = 𝑆 → ((( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅) ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆)))
9		dfrefrel2 39277	. 2 ⊢ ( RefRel 𝑅 ↔ (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅))
10		dfrefrel2 39277	. 2 ⊢ ( RefRel 𝑆 ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆))
11	8, 9, 10	3bitr4g 317	1 ⊢ (𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∩ cin 3905   ⊆ wss 3906   I cid 5557   × cxp 5661  dom cdm 5663  ran crn 5664  Rel wrel 5668   RefRel wrefrel 38871

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

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 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670  df-cnv 5671  df-dm 5673  df-rn 5674  df-res 5675  df-refrel 39274

This theorem is used by:  eqvreleq  39368