Theorem refreleq 36735

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 5825	. . . . . 6 ⊢ (𝑅 = 𝑆 → dom 𝑅 = dom 𝑆)
2		rneq 5857	. . . . . 6 ⊢ (𝑅 = 𝑆 → ran 𝑅 = ran 𝑆)
3	1, 2	xpeq12d 5631	. . . . 5 ⊢ (𝑅 = 𝑆 → (dom 𝑅 × ran 𝑅) = (dom 𝑆 × ran 𝑆))
4	3	ineq2d 4152	. . . 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 3959	. . 3 ⊢ (𝑅 = 𝑆 → (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ↔ ( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆))
7		releq 5698	. . 3 ⊢ (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
8	6, 7	anbi12d 632	. 2 ⊢ (𝑅 = 𝑆 → ((( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅) ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆)))
9		dfrefrel2 36729	. 2 ⊢ ( RefRel 𝑅 ↔ (( I ∩ (dom 𝑅 × ran 𝑅)) ⊆ 𝑅 ∧ Rel 𝑅))
10		dfrefrel2 36729	. 2 ⊢ ( RefRel 𝑆 ↔ (( I ∩ (dom 𝑆 × ran 𝑆)) ⊆ 𝑆 ∧ Rel 𝑆))
11	8, 9, 10	3bitr4g 314	1 ⊢ (𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 397   = wceq 1539   ∩ cin 3891   ⊆ wss 3892   I cid 5499   × cxp 5598  dom cdm 5600  ran crn 5601  Rel wrel 5605   RefRel wrefrel 36387

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-12 2169  ax-ext 2707  ax-sep 5232  ax-nul 5239  ax-pr 5361

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-3an 1089  df-tru 1542  df-fal 1552  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2714  df-cleq 2728  df-clel 2814  df-ral 3063  df-rex 3072  df-rab 3306  df-v 3439  df-dif 3895  df-un 3897  df-in 3899  df-ss 3909  df-nul 4263  df-if 4466  df-sn 4566  df-pr 4568  df-op 4572  df-br 5082  df-opab 5144  df-xp 5606  df-rel 5607  df-cnv 5608  df-dm 5610  df-rn 5611  df-res 5612  df-refrel 36726

This theorem is referenced by:  eqvreleq  36816