Theorem nnsucsssuc 6293

Description: Membership is inherited by successors. The reverse direction holds for all ordinals, as seen at onsucsssucr 4354, but the forward direction, for all ordinals, implies excluded middle as seen as onsucsssucexmid 4371. (Contributed by Jim Kingdon, 25-Aug-2019.)

Assertion

Ref	Expression
nnsucsssuc	⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))

Proof of Theorem nnsucsssuc

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

Step	Hyp	Ref	Expression
1		sseq1 3062	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
2		suceq 4253	. . . . . . 7 ⊢ (𝑥 = 𝐴 → suc 𝑥 = suc 𝐴)
3	2	sseq1d 3068	. . . . . 6 ⊢ (𝑥 = 𝐴 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))
4	1, 3	imbi12d 233	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵)))
5	4	imbi2d 229	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝐵 ∈ ω → (𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵)) ↔ (𝐵 ∈ ω → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵))))
6		sseq1 3062	. . . . . 6 ⊢ (𝑥 = ∅ → (𝑥 ⊆ 𝐵 ↔ ∅ ⊆ 𝐵))
7		suceq 4253	. . . . . . 7 ⊢ (𝑥 = ∅ → suc 𝑥 = suc ∅)
8	7	sseq1d 3068	. . . . . 6 ⊢ (𝑥 = ∅ → (suc 𝑥 ⊆ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
9	6, 8	imbi12d 233	. . . . 5 ⊢ (𝑥 = ∅ → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (∅ ⊆ 𝐵 → suc ∅ ⊆ suc 𝐵)))
10		sseq1 3062	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝑥 ⊆ 𝐵 ↔ 𝑦 ⊆ 𝐵))
11		suceq 4253	. . . . . . 7 ⊢ (𝑥 = 𝑦 → suc 𝑥 = suc 𝑦)
12	11	sseq1d 3068	. . . . . 6 ⊢ (𝑥 = 𝑦 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc 𝑦 ⊆ suc 𝐵))
13	10, 12	imbi12d 233	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵)))
14		sseq1 3062	. . . . . 6 ⊢ (𝑥 = suc 𝑦 → (𝑥 ⊆ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
15		suceq 4253	. . . . . . 7 ⊢ (𝑥 = suc 𝑦 → suc 𝑥 = suc suc 𝑦)
16	15	sseq1d 3068	. . . . . 6 ⊢ (𝑥 = suc 𝑦 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
17	14, 16	imbi12d 233	. . . . 5 ⊢ (𝑥 = suc 𝑦 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵)))
18		peano3 4439	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → suc 𝐵 ≠ ∅)
19	18	neneqd 2283	. . . . . . . 8 ⊢ (𝐵 ∈ ω → ¬ suc 𝐵 = ∅)
20		peano2 4438	. . . . . . . . . 10 ⊢ (𝐵 ∈ ω → suc 𝐵 ∈ ω)
21		0elnn 4460	. . . . . . . . . 10 ⊢ (suc 𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
22	20, 21	syl 14	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
23	22	ord 681	. . . . . . . 8 ⊢ (𝐵 ∈ ω → (¬ suc 𝐵 = ∅ → ∅ ∈ suc 𝐵))
24	19, 23	mpd 13	. . . . . . 7 ⊢ (𝐵 ∈ ω → ∅ ∈ suc 𝐵)
25		nnord 4454	. . . . . . . 8 ⊢ (𝐵 ∈ ω → Ord 𝐵)
26		ordsucim 4345	. . . . . . . 8 ⊢ (Ord 𝐵 → Ord suc 𝐵)
27		0ex 3987	. . . . . . . . 9 ⊢ ∅ ∈ V
28		ordelsuc 4350	. . . . . . . . 9 ⊢ ((∅ ∈ V ∧ Ord suc 𝐵) → (∅ ∈ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
29	27, 28	mpan 416	. . . . . . . 8 ⊢ (Ord suc 𝐵 → (∅ ∈ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
30	25, 26, 29	3syl 17	. . . . . . 7 ⊢ (𝐵 ∈ ω → (∅ ∈ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
31	24, 30	mpbid 146	. . . . . 6 ⊢ (𝐵 ∈ ω → suc ∅ ⊆ suc 𝐵)
32	31	a1d 22	. . . . 5 ⊢ (𝐵 ∈ ω → (∅ ⊆ 𝐵 → suc ∅ ⊆ suc 𝐵))
33		simp3 948	. . . . . . . . . 10 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ⊆ 𝐵)
34		simp1l 970	. . . . . . . . . . 11 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝑦 ∈ ω)
35		simp1r 971	. . . . . . . . . . . 12 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝐵 ∈ ω)
36	35, 25	syl 14	. . . . . . . . . . 11 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → Ord 𝐵)
37		ordelsuc 4350	. . . . . . . . . . 11 ⊢ ((𝑦 ∈ ω ∧ Ord 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
38	34, 36, 37	syl2anc 404	. . . . . . . . . 10 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
39	33, 38	mpbird 166	. . . . . . . . 9 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝑦 ∈ 𝐵)
40		nnsucelsuc 6292	. . . . . . . . . 10 ⊢ (𝐵 ∈ ω → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ∈ suc 𝐵))
41	35, 40	syl 14	. . . . . . . . 9 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ∈ suc 𝐵))
42	39, 41	mpbid 146	. . . . . . . 8 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ∈ suc 𝐵)
43		peano2 4438	. . . . . . . . . 10 ⊢ (𝑦 ∈ ω → suc 𝑦 ∈ ω)
44	34, 43	syl 14	. . . . . . . . 9 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ∈ ω)
45	36, 26	syl 14	. . . . . . . . 9 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → Ord suc 𝐵)
46		ordelsuc 4350	. . . . . . . . 9 ⊢ ((suc 𝑦 ∈ ω ∧ Ord suc 𝐵) → (suc 𝑦 ∈ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
47	44, 45, 46	syl2anc 404	. . . . . . . 8 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → (suc 𝑦 ∈ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
48	42, 47	mpbid 146	. . . . . . 7 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc suc 𝑦 ⊆ suc 𝐵)
49	48	3expia 1148	. . . . . 6 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵)) → (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵))
50	49	exp31 357	. . . . 5 ⊢ (𝑦 ∈ ω → (𝐵 ∈ ω → ((𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) → (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵))))
51	9, 13, 17, 32, 50	finds2 4444	. . . 4 ⊢ (𝑥 ∈ ω → (𝐵 ∈ ω → (𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵)))
52	5, 51	vtoclga 2699	. . 3 ⊢ (𝐴 ∈ ω → (𝐵 ∈ ω → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵)))
53	52	imp 123	. 2 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵))
54		nnon 4452	. . 3 ⊢ (𝐴 ∈ ω → 𝐴 ∈ On)
55		onsucsssucr 4354	. . 3 ⊢ ((𝐴 ∈ On ∧ Ord 𝐵) → (suc 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵))
56	54, 25, 55	syl2an 284	. 2 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵))
57	53, 56	impbid 128	1 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 667   ∧ w3a 927   = wceq 1296   ∈ wcel 1445  Vcvv 2633   ⊆ wss 3013  ∅c0 3302  Ord word 4213  Oncon0 4214  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-sep 3978  ax-nul 3986  ax-pow 4030  ax-pr 4060  ax-un 4284  ax-iinf 4431

This theorem depends on definitions:  df-bi 116  df-3an 929  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-ne 2263  df-ral 2375  df-rex 2376  df-v 2635  df-dif 3015  df-un 3017  df-in 3019  df-ss 3026  df-nul 3303  df-pw 3451  df-sn 3472  df-pr 3473  df-uni 3676  df-int 3711  df-tr 3959  df-iord 4217  df-on 4219  df-suc 4222  df-iom 4434

This theorem is referenced by:  nnaword  6310