Theorem bj-nn0suc0 13985

Description: Constructive proof of a variant of nn0suc 4588. For a constructive proof of nn0suc 4588, see bj-nn0suc 13999. (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 2177	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 = ∅ ↔ 𝐴 = ∅))
2		eqeq1 2177	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 = suc 𝑥 ↔ 𝐴 = suc 𝑥))
3	2	rexeqbi1dv 2674	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥 ↔ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))
4	1, 3	orbi12d 788	. 2 ⊢ (𝑦 = 𝐴 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) ↔ (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥)))
5		tru 1352	. . 3 ⊢ ⊤
6		a1tru 1364	. . . 4 ⊢ (⊤ → ⊤)
7	6	rgenw 2525	. . 3 ⊢ ∀𝑧 ∈ ω (⊤ → ⊤)
8		bdeq0 13902	. . . . 5 ⊢ BOUNDED 𝑦 = ∅
9		bdeqsuc 13916	. . . . . 6 ⊢ BOUNDED 𝑦 = suc 𝑥
10	9	ax-bdex 13854	. . . . 5 ⊢ BOUNDED ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥
11	8, 10	ax-bdor 13851	. . . 4 ⊢ BOUNDED (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
12		nfv 1521	. . . 4 ⊢ Ⅎ𝑦⊤
13		orc 707	. . . . 5 ⊢ (𝑦 = ∅ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
14	13	a1d 22	. . . 4 ⊢ (𝑦 = ∅ → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
15		a1tru 1364	. . . . 5 ⊢ (¬ (𝑦 = 𝑧 → ¬ (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)) → ⊤)
16	15	expi 633	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) → ⊤))
17		vex 2733	. . . . . . . . 9 ⊢ 𝑧 ∈ V
18	17	sucid 4402	. . . . . . . 8 ⊢ 𝑧 ∈ suc 𝑧
19		eleq2 2234	. . . . . . . 8 ⊢ (𝑦 = suc 𝑧 → (𝑧 ∈ 𝑦 ↔ 𝑧 ∈ suc 𝑧))
20	18, 19	mpbiri 167	. . . . . . 7 ⊢ (𝑦 = suc 𝑧 → 𝑧 ∈ 𝑦)
21		suceq 4387	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → suc 𝑥 = suc 𝑧)
22	21	eqeq2d 2182	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → (𝑦 = suc 𝑥 ↔ 𝑦 = suc 𝑧))
23	22	rspcev 2834	. . . . . . 7 ⊢ ((𝑧 ∈ 𝑦 ∧ 𝑦 = suc 𝑧) → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
24	20, 23	mpancom 420	. . . . . 6 ⊢ (𝑦 = suc 𝑧 → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
25	24	olcd 729	. . . . 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 13982	. . 3 ⊢ ((⊤ ∧ ∀𝑧 ∈ ω (⊤ → ⊤)) → ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
28	5, 7, 27	mp2an 424	. 2 ⊢ ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
29	4, 28	vtoclri 2805	1 ⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 703   = wceq 1348  ⊤wtru 1349   ∈ wcel 2141  ∀wral 2448  ∃wrex 2449  ∅c0 3414  suc csuc 4350  ωcom 4574

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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-13 2143  ax-14 2144  ax-ext 2152  ax-nul 4115  ax-pr 4194  ax-un 4418  ax-bd0 13848  ax-bdim 13849  ax-bdan 13850  ax-bdor 13851  ax-bdn 13852  ax-bdal 13853  ax-bdex 13854  ax-bdeq 13855  ax-bdel 13856  ax-bdsb 13857  ax-bdsep 13919  ax-infvn 13976

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-fal 1354  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-rab 2457  df-v 2732  df-dif 3123  df-un 3125  df-in 3127  df-ss 3134  df-nul 3415  df-sn 3589  df-pr 3590  df-uni 3797  df-int 3832  df-suc 4356  df-iom 4575  df-bdc 13876  df-bj-ind 13962

This theorem is referenced by:  bj-nn0suc  13999