Theorem bj-nn0suc0 12553

Description: Constructive proof of a variant of nn0suc 4447. For a constructive proof of nn0suc 4447, see bj-nn0suc 12567. (Contributed by BJ, 19-Nov-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-nn0suc0	⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Distinct variable group:   𝑥,𝐴

Proof of Theorem bj-nn0suc0

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

Step	Hyp	Ref	Expression
1		eqeq1 2101	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 = ∅ ↔ 𝐴 = ∅))
2		eqeq1 2101	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 = suc 𝑥 ↔ 𝐴 = suc 𝑥))
3	2	rexeqbi1dv 2585	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥 ↔ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))
4	1, 3	orbi12d 745	. 2 ⊢ (𝑦 = 𝐴 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) ↔ (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥)))
5		tru 1300	. . 3 ⊢ ⊤
6		a1tru 1312	. . . 4 ⊢ (⊤ → ⊤)
7	6	rgenw 2441	. . 3 ⊢ ∀𝑧 ∈ ω (⊤ → ⊤)
8		bdeq0 12466	. . . . 5 ⊢ BOUNDED 𝑦 = ∅
9		bdeqsuc 12480	. . . . . 6 ⊢ BOUNDED 𝑦 = suc 𝑥
10	9	ax-bdex 12418	. . . . 5 ⊢ BOUNDED ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥
11	8, 10	ax-bdor 12415	. . . 4 ⊢ BOUNDED (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
12		nfv 1473	. . . 4 ⊢ Ⅎ𝑦⊤
13		orc 671	. . . . 5 ⊢ (𝑦 = ∅ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
14	13	a1d 22	. . . 4 ⊢ (𝑦 = ∅ → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
15		a1tru 1312	. . . . 5 ⊢ (¬ (𝑦 = 𝑧 → ¬ (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)) → ⊤)
16	15	expi 605	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) → ⊤))
17		vex 2636	. . . . . . . . 9 ⊢ 𝑧 ∈ V
18	17	sucid 4268	. . . . . . . 8 ⊢ 𝑧 ∈ suc 𝑧
19		eleq2 2158	. . . . . . . 8 ⊢ (𝑦 = suc 𝑧 → (𝑧 ∈ 𝑦 ↔ 𝑧 ∈ suc 𝑧))
20	18, 19	mpbiri 167	. . . . . . 7 ⊢ (𝑦 = suc 𝑧 → 𝑧 ∈ 𝑦)
21		suceq 4253	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → suc 𝑥 = suc 𝑧)
22	21	eqeq2d 2106	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → (𝑦 = suc 𝑥 ↔ 𝑦 = suc 𝑧))
23	22	rspcev 2736	. . . . . . 7 ⊢ ((𝑧 ∈ 𝑦 ∧ 𝑦 = suc 𝑧) → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
24	20, 23	mpancom 414	. . . . . 6 ⊢ (𝑦 = suc 𝑧 → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
25	24	olcd 691	. . . . 5 ⊢ (𝑦 = suc 𝑧 → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
26	25	a1d 22	. . . 4 ⊢ (𝑦 = suc 𝑧 → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
27	11, 12, 12, 12, 14, 16, 26	bj-bdfindis 12550	. . 3 ⊢ ((⊤ ∧ ∀𝑧 ∈ ω (⊤ → ⊤)) → ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
28	5, 7, 27	mp2an 418	. 2 ⊢ ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
29	4, 28	vtoclri 2708	1 ⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 667   = wceq 1296  ⊤wtru 1297   ∈ wcel 1445  ∀wral 2370  ∃wrex 2371  ∅c0 3302  suc csuc 4216  ωcom 4433

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-13 1456  ax-14 1457  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077  ax-nul 3986  ax-pr 4060  ax-un 4284  ax-bd0 12412  ax-bdim 12413  ax-bdan 12414  ax-bdor 12415  ax-bdn 12416  ax-bdal 12417  ax-bdex 12418  ax-bdeq 12419  ax-bdel 12420  ax-bdsb 12421  ax-bdsep 12483  ax-infvn 12544

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-fal 1302  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-ral 2375  df-rex 2376  df-rab 2379  df-v 2635  df-dif 3015  df-un 3017  df-in 3019  df-ss 3026  df-nul 3303  df-sn 3472  df-pr 3473  df-uni 3676  df-int 3711  df-suc 4222  df-iom 4434  df-bdc 12440  df-bj-ind 12530

This theorem is referenced by:  bj-nn0suc  12567