Theorem bj-nn0suc0 13319

Description: Constructive proof of a variant of nn0suc 4526. For a constructive proof of nn0suc 4526, see bj-nn0suc 13333. (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 2147	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 = ∅ ↔ 𝐴 = ∅))
2		eqeq1 2147	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 = suc 𝑥 ↔ 𝐴 = suc 𝑥))
3	2	rexeqbi1dv 2638	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥 ↔ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))
4	1, 3	orbi12d 783	. 2 ⊢ (𝑦 = 𝐴 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) ↔ (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥)))
5		tru 1336	. . 3 ⊢ ⊤
6		a1tru 1348	. . . 4 ⊢ (⊤ → ⊤)
7	6	rgenw 2490	. . 3 ⊢ ∀𝑧 ∈ ω (⊤ → ⊤)
8		bdeq0 13236	. . . . 5 ⊢ BOUNDED 𝑦 = ∅
9		bdeqsuc 13250	. . . . . 6 ⊢ BOUNDED 𝑦 = suc 𝑥
10	9	ax-bdex 13188	. . . . 5 ⊢ BOUNDED ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥
11	8, 10	ax-bdor 13185	. . . 4 ⊢ BOUNDED (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
12		nfv 1509	. . . 4 ⊢ Ⅎ𝑦⊤
13		orc 702	. . . . 5 ⊢ (𝑦 = ∅ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
14	13	a1d 22	. . . 4 ⊢ (𝑦 = ∅ → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
15		a1tru 1348	. . . . 5 ⊢ (¬ (𝑦 = 𝑧 → ¬ (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)) → ⊤)
16	15	expi 628	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) → ⊤))
17		vex 2692	. . . . . . . . 9 ⊢ 𝑧 ∈ V
18	17	sucid 4347	. . . . . . . 8 ⊢ 𝑧 ∈ suc 𝑧
19		eleq2 2204	. . . . . . . 8 ⊢ (𝑦 = suc 𝑧 → (𝑧 ∈ 𝑦 ↔ 𝑧 ∈ suc 𝑧))
20	18, 19	mpbiri 167	. . . . . . 7 ⊢ (𝑦 = suc 𝑧 → 𝑧 ∈ 𝑦)
21		suceq 4332	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → suc 𝑥 = suc 𝑧)
22	21	eqeq2d 2152	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → (𝑦 = suc 𝑥 ↔ 𝑦 = suc 𝑧))
23	22	rspcev 2793	. . . . . . 7 ⊢ ((𝑧 ∈ 𝑦 ∧ 𝑦 = suc 𝑧) → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
24	20, 23	mpancom 419	. . . . . 6 ⊢ (𝑦 = suc 𝑧 → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
25	24	olcd 724	. . . . 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 13316	. . 3 ⊢ ((⊤ ∧ ∀𝑧 ∈ ω (⊤ → ⊤)) → ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
28	5, 7, 27	mp2an 423	. 2 ⊢ ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
29	4, 28	vtoclri 2764	1 ⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 698   = wceq 1332  ⊤wtru 1333   ∈ wcel 1481  ∀wral 2417  ∃wrex 2418  ∅c0 3368  suc csuc 4295  ωcom 4512

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-13 1492  ax-14 1493  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-nul 4062  ax-pr 4139  ax-un 4363  ax-bd0 13182  ax-bdim 13183  ax-bdan 13184  ax-bdor 13185  ax-bdn 13186  ax-bdal 13187  ax-bdex 13188  ax-bdeq 13189  ax-bdel 13190  ax-bdsb 13191  ax-bdsep 13253  ax-infvn 13310

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-fal 1338  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-rab 2426  df-v 2691  df-dif 3078  df-un 3080  df-in 3082  df-ss 3089  df-nul 3369  df-sn 3538  df-pr 3539  df-uni 3745  df-int 3780  df-suc 4301  df-iom 4513  df-bdc 13210  df-bj-ind 13296

This theorem is referenced by:  bj-nn0suc  13333