Theorem relexp0eq 44455

Description: The zeroth power of relationships is the same if and only if the union of their domain and ranges is the same. (Contributed by RP, 11-Jun-2020.)

Assertion

Ref	Expression
relexp0eq	⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) → ((dom 𝐴 ∪ ran 𝐴) = (dom 𝐵 ∪ ran 𝐵) ↔ (𝐴↑𝑟0) = (𝐵↑𝑟0)))

Proof of Theorem relexp0eq

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2755	. . . 4 ⊢ ((dom 𝐴 ∪ ran 𝐴) = (dom 𝐵 ∪ ran 𝐵) ↔ ∀𝑥(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)))
2		alcom 2193	. . . . 5 ⊢ (∀𝑦∀𝑥(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑥∀𝑦(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)))
3		19.3v 2011	. . . . 5 ⊢ (∀𝑦∀𝑥(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑥(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)))
4		ax6ev 1998	. . . . . . . . 9 ⊢ ∃𝑦 𝑦 = 𝑥
5		pm5.5 364	. . . . . . . . 9 ⊢ (∃𝑦 𝑦 = 𝑥 → ((∃𝑦 𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))))
6	4, 5	ax-mp 5	. . . . . . . 8 ⊢ ((∃𝑦 𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)))
7		19.23v 1971	. . . . . . . 8 ⊢ (∀𝑦(𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ (∃𝑦 𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))))
8		19.3v 2011	. . . . . . . 8 ⊢ (∀𝑦(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)))
9	6, 7, 8	3bitr4ri 307	. . . . . . 7 ⊢ (∀𝑦(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑦(𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))))
10		pm5.32 583	. . . . . . . . 9 ⊢ ((𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ ((𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐴 ∪ ran 𝐴)) ↔ (𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))))
11		ancom 465	. . . . . . . . . 10 ⊢ ((𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐴 ∪ ran 𝐴)) ↔ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥))
12		ancom 465	. . . . . . . . . 10 ⊢ ((𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥))
13	11, 12	bibi12i 342	. . . . . . . . 9 ⊢ (((𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐴 ∪ ran 𝐴)) ↔ (𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ ((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
14	10, 13	bitri 278	. . . . . . . 8 ⊢ ((𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ ((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
15	14	albii 1848	. . . . . . 7 ⊢ (∀𝑦(𝑦 = 𝑥 → (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵))) ↔ ∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
16	9, 15	bitri 278	. . . . . 6 ⊢ (∀𝑦(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
17	16	albii 1848	. . . . 5 ⊢ (∀𝑥∀𝑦(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑥∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
18	2, 3, 17	3bitr3i 304	. . . 4 ⊢ (∀𝑥(𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ↔ 𝑥 ∈ (dom 𝐵 ∪ ran 𝐵)) ↔ ∀𝑥∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
19	1, 18	bitri 278	. . 3 ⊢ ((dom 𝐴 ∪ ran 𝐴) = (dom 𝐵 ∪ ran 𝐵) ↔ ∀𝑥∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
20		eqopab2bw 5532	. . 3 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)} ↔ ∀𝑥∀𝑦((𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥) ↔ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)))
21		opabresid 6051	. . . . 5 ⊢ ( I ↾ (dom 𝐴 ∪ ran 𝐴)) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)}
22	21	eqcomi 2771	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)} = ( I ↾ (dom 𝐴 ∪ ran 𝐴))
23		opabresid 6051	. . . . 5 ⊢ ( I ↾ (dom 𝐵 ∪ ran 𝐵)) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)}
24	23	eqcomi 2771	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)} = ( I ↾ (dom 𝐵 ∪ ran 𝐵))
25	22, 24	eqeq12i 2780	. . 3 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)} ↔ ( I ↾ (dom 𝐴 ∪ ran 𝐴)) = ( I ↾ (dom 𝐵 ∪ ran 𝐵)))
26	19, 20, 25	3bitr2i 302	. 2 ⊢ ((dom 𝐴 ∪ ran 𝐴) = (dom 𝐵 ∪ ran 𝐵) ↔ ( I ↾ (dom 𝐴 ∪ ran 𝐴)) = ( I ↾ (dom 𝐵 ∪ ran 𝐵)))
27		relexp0g 15066	. . 3 ⊢ (𝐴 ∈ 𝑈 → (𝐴↑𝑟0) = ( I ↾ (dom 𝐴 ∪ ran 𝐴)))
28		relexp0g 15066	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐵↑𝑟0) = ( I ↾ (dom 𝐵 ∪ ran 𝐵)))
29	27, 28	eqeqan12d 2776	. 2 ⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) → ((𝐴↑𝑟0) = (𝐵↑𝑟0) ↔ ( I ↾ (dom 𝐴 ∪ ran 𝐴)) = ( I ↾ (dom 𝐵 ∪ ran 𝐵))))
30	26, 29	bitr4id 293	1 ⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) → ((dom 𝐴 ∪ ran 𝐴) = (dom 𝐵 ∪ ran 𝐵) ↔ (𝐴↑𝑟0) = (𝐵↑𝑟0)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1567   = wceq 1569  ∃wex 1808   ∈ wcel 2142   ∪ cun 3902  {copab 5172   I cid 5554  dom cdm 5660  ran crn 5661   ↾ cres 5662  (class class class)co 7412  0cc0 11106  ↑_𝑟crelexp 15063

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734  ax-sep 5256  ax-pow 5335  ax-pr 5403  ax-un 7734  ax-1cn 11164  ax-icn 11165  ax-addcl 11166  ax-mulcl 11168  ax-i2m1 11174

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-nf 1813  df-sb 2096  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ral 3079  df-rex 3089  df-rab 3416  df-v 3456  df-sbc 3744  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-pw 4563  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-opab 5173  df-id 5555  df-xp 5666  df-rel 5667  df-cnv 5668  df-co 5669  df-dm 5670  df-rn 5671  df-res 5672  df-iota 6492  df-fun 6538  df-fv 6544  df-ov 7415  df-oprab 7416  df-mpo 7417  df-n0 12511  df-relexp 15064

This theorem is used by:  iunrelexp0  44456