Theorem relexp0eq 44639

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 2753	. . . 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 5520	. . 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 6041	. . . . 5 ⊢ ( I ↾ (dom 𝐴 ∪ ran 𝐴)) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)}
22	21	eqcomi 2769	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐴 ∪ ran 𝐴) ∧ 𝑦 = 𝑥)} = ( I ↾ (dom 𝐴 ∪ ran 𝐴))
23		opabresid 6041	. . . . 5 ⊢ ( I ↾ (dom 𝐵 ∪ ran 𝐵)) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)}
24	23	eqcomi 2769	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ (dom 𝐵 ∪ ran 𝐵) ∧ 𝑦 = 𝑥)} = ( I ↾ (dom 𝐵 ∪ ran 𝐵))
25	22, 24	eqeq12i 2778	. . 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 15128	. . 3 ⊢ (𝐴 ∈ 𝑈 → (𝐴↑𝑟0) = ( I ↾ (dom 𝐴 ∪ ran 𝐴)))
28		relexp0g 15128	. . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐵↑𝑟0) = ( I ↾ (dom 𝐵 ∪ ran 𝐵)))
29	27, 28	eqeqan12d 2774	. 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 3897  {copab 5167   I cid 5542  dom cdm 5648  ran crn 5649   ↾ cres 5650  (class class class)co 7409  0cc0 11157  ↑_𝑟crelexp 15125

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 2213  ax-ext 2732  ax-sep 5249  ax-pow 5327  ax-pr 5391  ax-un 7735  ax-1cn 11215  ax-icn 11216  ax-addcl 11217  ax-mulcl 11219  ax-i2m1 11225

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 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-sbc 3740  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-id 5543  df-xp 5654  df-rel 5655  df-cnv 5656  df-co 5657  df-dm 5658  df-rn 5659  df-res 5660  df-iota 6484  df-fun 6530  df-fv 6536  df-ov 7412  df-oprab 7413  df-mpo 7414  df-n0 12562  df-relexp 15126

This theorem is used by:  iunrelexp0  44640