Theorem refreleq 39640

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 5885	. . . . . 6 ⊢ (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
2		rneq 5918	. . . . . 6 ⊢ (𝑅 = 𝑆 → ran 𝑅 = ran 𝑆)
3	1, 2	xpeq12d 5681	. . . . 5 ⊢ (𝑅 = 𝑆 → (dom 𝑅 × ran 𝑅) = (dom 𝑆 × ran 𝑆))
4	3	ineq2d 4116	. . . 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 3942	. . 3 ⊢ (𝑅 = 𝑆 → (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ↔ ( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆))
7		releq 5753	. . 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 39634	. 2 ⊢ ( RefRel 𝑅 ↔ (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅))
10		dfrefrel2 39634	. 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 1569   ⊆ wss 3896   ∩ cin 4088   I cid 5544   × cxp 5648  dom cdm 5650  ran crn 5651  Rel wrel 5655   RefRel wrefrel 39228

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-xp 5656  df-rel 5657  df-cnv 5658  df-dm 5660  df-rn 5661  df-res 5662  df-refrel 39631

This theorem is used by:  eqvreleq  39725