Theorem relexp0eq 44528

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 2196	. . . . 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 2015	. . . . 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 2002	. . . . . . . . 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 1975	. . . . . . . 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 2015	. . . . . . . 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		pm5.32 584	. . . . . . . . 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 466	. . . . . . . . . 10 ⊢ ((𝑦 = 𝑥 ∧ 𝑥 ∈ (dom 𝐴 ∪ ran 𝐴)) ↔ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥))
12		ancom 466	. . . . . . . . . 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 1852	. . . . . . 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 1852	. . . . 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 5531	. . 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 6050	. . . . 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 6050	. . . . 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 15097	. . 3 ⊢ (𝐴 ∈ 𝑈 → (𝐴↑𝑟0) = ( I ↾ (dom 𝐴 ∪ ran 𝐴)))
28		relexp0g 15097	. . 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 401  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145   ∪ cun 3900  {copab 5171   I cid 5553  dom cdm 5659  ran crn 5660   ↾ cres 5661  (class class class)co 7416  0cc0 11127  ↑_𝑟crelexp 15094

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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734  ax-sep 5255  ax-pow 5334  ax-pr 5402  ax-un 7739  ax-1cn 11185  ax-icn 11186  ax-addcl 11187  ax-mulcl 11189  ax-i2m1 11195

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-nf 1817  df-sb 2100  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 3415  df-v 3455  df-sbc 3743  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-pw 4562  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-id 5554  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-res 5671  df-iota 6493  df-fun 6539  df-fv 6545  df-ov 7419  df-oprab 7420  df-mpo 7421  df-n0 12532  df-relexp 15095

This theorem is used by:  iunrelexp0  44529