Theorem bj-nn0suc0 10075

Description: Constructive proof of a variant of nn0suc 4327. For a constructive proof of nn0suc 4327, see bj-nn0suc 10089. (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 2046	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 = ∅ ↔ 𝐴 = ∅))
2		eqeq1 2046	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 = suc 𝑥 ↔ 𝐴 = suc 𝑥))
3	2	rexeqbi1dv 2514	. . 3 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥 ↔ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))
4	1, 3	orbi12d 707	. 2 ⊢ (𝑦 = 𝐴 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) ↔ (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥)))
5		tru 1247	. . 3 ⊢ ⊤
6		a1tru 1259	. . . 4 ⊢ (⊤ → ⊤)
7	6	rgenw 2376	. . 3 ⊢ ∀𝑧 ∈ ω (⊤ → ⊤)
8		bdeq0 9987	. . . . 5 ⊢ BOUNDED 𝑦 = ∅
9		bdeqsuc 10001	. . . . . 6 ⊢ BOUNDED 𝑦 = suc 𝑥
10	9	ax-bdex 9939	. . . . 5 ⊢ BOUNDED ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥
11	8, 10	ax-bdor 9936	. . . 4 ⊢ BOUNDED (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
12		nfv 1421	. . . 4 ⊢ Ⅎ𝑦⊤
13		orc 633	. . . . 5 ⊢ (𝑦 = ∅ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
14	13	a1d 22	. . . 4 ⊢ (𝑦 = ∅ → (⊤ → (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)))
15		a1tru 1259	. . . . 5 ⊢ (¬ (𝑦 = 𝑧 → ¬ (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)) → ⊤)
16	15	expi 567	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥) → ⊤))
17		vex 2560	. . . . . . . . 9 ⊢ 𝑧 ∈ V
18	17	sucid 4154	. . . . . . . 8 ⊢ 𝑧 ∈ suc 𝑧
19		eleq2 2101	. . . . . . . 8 ⊢ (𝑦 = suc 𝑧 → (𝑧 ∈ 𝑦 ↔ 𝑧 ∈ suc 𝑧))
20	18, 19	mpbiri 157	. . . . . . 7 ⊢ (𝑦 = suc 𝑧 → 𝑧 ∈ 𝑦)
21		suceq 4139	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → suc 𝑥 = suc 𝑧)
22	21	eqeq2d 2051	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → (𝑦 = suc 𝑥 ↔ 𝑦 = suc 𝑧))
23	22	rspcev 2656	. . . . . . 7 ⊢ ((𝑧 ∈ 𝑦 ∧ 𝑦 = suc 𝑧) → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
24	20, 23	mpancom 399	. . . . . 6 ⊢ (𝑦 = suc 𝑧 → ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
25	24	olcd 653	. . . . 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 10072	. . 3 ⊢ ((⊤ ∧ ∀𝑧 ∈ ω (⊤ → ⊤)) → ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥))
28	5, 7, 27	mp2an 402	. 2 ⊢ ∀𝑦 ∈ ω (𝑦 = ∅ ∨ ∃𝑥 ∈ 𝑦 𝑦 = suc 𝑥)
29	4, 28	vtoclri 2628	1 ⊢ (𝐴 ∈ ω → (𝐴 = ∅ ∨ ∃𝑥 ∈ 𝐴 𝐴 = suc 𝑥))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 629   = wceq 1243  ⊤wtru 1244   ∈ wcel 1393  ∀wral 2306  ∃wrex 2307  ∅c0 3224  suc csuc 4102  ωcom 4313

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-13 1404  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-nul 3883  ax-pr 3944  ax-un 4170  ax-bd0 9933  ax-bdim 9934  ax-bdan 9935  ax-bdor 9936  ax-bdn 9937  ax-bdal 9938  ax-bdex 9939  ax-bdeq 9940  ax-bdel 9941  ax-bdsb 9942  ax-bdsep 10004  ax-infvn 10066

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-fal 1249  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-rab 2315  df-v 2559  df-dif 2920  df-un 2922  df-in 2924  df-ss 2931  df-nul 3225  df-sn 3381  df-pr 3382  df-uni 3581  df-int 3616  df-suc 4108  df-iom 4314  df-bdc 9961  df-bj-ind 10051

This theorem is referenced by:  bj-nn0suc  10089