Theorem bj-nn0suc0 15512

Description: Constructive proof of a variant of nn0suc 4637. For a constructive proof of nn0suc 4637, see bj-nn0suc 15526. (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 2200	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 = ∅ ↔ 𝐴 = ∅))
2		eqeq1 2200	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 = suc 𝑥 ↔ 𝐴 = suc 𝑥))
3	2	rexeqbi1dv 2703	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥 ↔ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))
4	1, 3	orbi12d 794	. 2 ⊢ (𝑦 = 𝐴 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) ↔ (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥)))
5		tru 1368	. . 3 ⊢ ⊤
6		trud 1380	. . . 4 ⊢ (⊤ → ⊤)
7	6	rgenw 2549	. . 3 ⊢ ∀𝑧 ∈ ω (⊤ → ⊤)
8		bdeq0 15429	. . . . 5 ⊢ BOUNDED 𝑦 = ∅
9		bdeqsuc 15443	. . . . . 6 ⊢ BOUNDED 𝑦 = suc 𝑥
10	9	ax-bdex 15381	. . . . 5 ⊢ BOUNDED ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥
11	8, 10	ax-bdor 15378	. . . 4 ⊢ BOUNDED (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
12		nfv 1539	. . . 4 ⊢ Ⅎ𝑦⊤
13		orc 713	. . . . 5 ⊢ (𝑦 = ∅ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
14	13	a1d 22	. . . 4 ⊢ (𝑦 = ∅ → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
15		trud 1380	. . . . 5 ⊢ (¬ (𝑦 = 𝑧 → ¬ (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)) → ⊤)
16	15	expi 639	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) → ⊤))
17		vex 2763	. . . . . . . . 9 ⊢ 𝑧 ∈ V
18	17	sucid 4449	. . . . . . . 8 ⊢ 𝑧 ∈ suc 𝑧
19		eleq2 2257	. . . . . . . 8 ⊢ (𝑦 = suc 𝑧 → (𝑧 ∈ 𝑦 ↔ 𝑧 ∈ suc 𝑧))
20	18, 19	mpbiri 168	. . . . . . 7 ⊢ (𝑦 = suc 𝑧 → 𝑧 ∈ 𝑦)
21		suceq 4434	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → suc 𝑥 = suc 𝑧)
22	21	eqeq2d 2205	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → (𝑦 = suc 𝑥 ↔ 𝑦 = suc 𝑧))
23	22	rspcev 2865	. . . . . . 7 ⊢ ((𝑧 ∈ 𝑦 ∧ 𝑦 = suc 𝑧) → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
24	20, 23	mpancom 422	. . . . . 6 ⊢ (𝑦 = suc 𝑧 → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
25	24	olcd 735	. . . . 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 15509	. . 3 ⊢ ((⊤ ∧ ∀𝑧 ∈ ω (⊤ → ⊤)) → ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
28	5, 7, 27	mp2an 426	. 2 ⊢ ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
29	4, 28	vtoclri 2836	1 ⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 709   = wceq 1364  ⊤wtru 1365   ∈ wcel 2164  ∀wral 2472  ∃wrex 2473  ∅c0 3447  suc csuc 4397  ωcom 4623

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-13 2166  ax-14 2167  ax-ext 2175  ax-nul 4156  ax-pr 4239  ax-un 4465  ax-bd0 15375  ax-bdim 15376  ax-bdan 15377  ax-bdor 15378  ax-bdn 15379  ax-bdal 15380  ax-bdex 15381  ax-bdeq 15382  ax-bdel 15383  ax-bdsb 15384  ax-bdsep 15446  ax-infvn 15503

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rex 2478  df-rab 2481  df-v 2762  df-dif 3156  df-un 3158  df-in 3160  df-ss 3167  df-nul 3448  df-sn 3625  df-pr 3626  df-uni 3837  df-int 3872  df-suc 4403  df-iom 4624  df-bdc 15403  df-bj-ind 15489

This theorem is referenced by:  bj-nn0suc  15526