Theorem nnsucsssuc 6188

Description: Membership is inherited by successors. The reverse direction holds for all ordinals, as seen at onsucsssucr 4292, but the forward direction, for all ordinals, implies excluded middle as seen as onsucsssucexmid 4309. (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 3033	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
2		suceq 4196	. . . . . . 7 ⊢ (𝑥 = 𝐴 → suc 𝑥 = suc 𝐴)
3	2	sseq1d 3039	. . . . . 6 ⊢ (𝑥 = 𝐴 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))
4	1, 3	imbi12d 232	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵)))
5	4	imbi2d 228	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝐵 ∈ ω → (𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵)) ↔ (𝐵 ∈ ω → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵))))
6		sseq1 3033	. . . . . 6 ⊢ (𝑥 = ∅ → (𝑥 ⊆ 𝐵 ↔ ∅ ⊆ 𝐵))
7		suceq 4196	. . . . . . 7 ⊢ (𝑥 = ∅ → suc 𝑥 = suc ∅)
8	7	sseq1d 3039	. . . . . 6 ⊢ (𝑥 = ∅ → (suc 𝑥 ⊆ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
9	6, 8	imbi12d 232	. . . . 5 ⊢ (𝑥 = ∅ → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (∅ ⊆ 𝐵 → suc ∅ ⊆ suc 𝐵)))
10		sseq1 3033	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝑥 ⊆ 𝐵 ↔ 𝑦 ⊆ 𝐵))
11		suceq 4196	. . . . . . 7 ⊢ (𝑥 = 𝑦 → suc 𝑥 = suc 𝑦)
12	11	sseq1d 3039	. . . . . 6 ⊢ (𝑥 = 𝑦 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc 𝑦 ⊆ suc 𝐵))
13	10, 12	imbi12d 232	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵)))
14		sseq1 3033	. . . . . 6 ⊢ (𝑥 = suc 𝑦 → (𝑥 ⊆ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
15		suceq 4196	. . . . . . 7 ⊢ (𝑥 = suc 𝑦 → suc 𝑥 = suc suc 𝑦)
16	15	sseq1d 3039	. . . . . 6 ⊢ (𝑥 = suc 𝑦 → (suc 𝑥 ⊆ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
17	14, 16	imbi12d 232	. . . . 5 ⊢ (𝑥 = suc 𝑦 → ((𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵) ↔ (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵)))
18		peano3 4377	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → suc 𝐵 ≠ ∅)
19	18	neneqd 2272	. . . . . . . 8 ⊢ (𝐵 ∈ ω → ¬ suc 𝐵 = ∅)
20		peano2 4376	. . . . . . . . . 10 ⊢ (𝐵 ∈ ω → suc 𝐵 ∈ ω)
21		0elnn 4398	. . . . . . . . . 10 ⊢ (suc 𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
22	20, 21	syl 14	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
23	22	ord 676	. . . . . . . 8 ⊢ (𝐵 ∈ ω → (¬ suc 𝐵 = ∅ → ∅ ∈ suc 𝐵))
24	19, 23	mpd 13	. . . . . . 7 ⊢ (𝐵 ∈ ω → ∅ ∈ suc 𝐵)
25		nnord 4392	. . . . . . . 8 ⊢ (𝐵 ∈ ω → Ord 𝐵)
26		ordsucim 4283	. . . . . . . 8 ⊢ (Ord 𝐵 → Ord suc 𝐵)
27		0ex 3934	. . . . . . . . 9 ⊢ ∅ ∈ V
28		ordelsuc 4288	. . . . . . . . 9 ⊢ ((∅ ∈ V ∧ Ord suc 𝐵) → (∅ ∈ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
29	27, 28	mpan 415	. . . . . . . 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 145	. . . . . 6 ⊢ (𝐵 ∈ ω → suc ∅ ⊆ suc 𝐵)
32	31	a1d 22	. . . . 5 ⊢ (𝐵 ∈ ω → (∅ ⊆ 𝐵 → suc ∅ ⊆ suc 𝐵))
33		simp3 943	. . . . . . . . . 10 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ⊆ 𝐵)
34		simp1l 965	. . . . . . . . . . 11 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝑦 ∈ ω)
35		simp1r 966	. . . . . . . . . . . 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 4288	. . . . . . . . . . 11 ⊢ ((𝑦 ∈ ω ∧ Ord 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
38	34, 36, 37	syl2anc 403	. . . . . . . . . 10 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
39	33, 38	mpbird 165	. . . . . . . . 9 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝑦 ∈ 𝐵)
40		nnsucelsuc 6187	. . . . . . . . . 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 145	. . . . . . . 8 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ∈ suc 𝐵)
43		peano2 4376	. . . . . . . . . 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 4288	. . . . . . . . 9 ⊢ ((suc 𝑦 ∈ ω ∧ Ord suc 𝐵) → (suc 𝑦 ∈ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
47	44, 45, 46	syl2anc 403	. . . . . . . 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 145	. . . . . . 7 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc suc 𝑦 ⊆ suc 𝐵)
49	48	3expia 1143	. . . . . 6 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵)) → (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵))
50	49	exp31 356	. . . . 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 4382	. . . 4 ⊢ (𝑥 ∈ ω → (𝐵 ∈ ω → (𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵)))
52	5, 51	vtoclga 2677	. . 3 ⊢ (𝐴 ∈ ω → (𝐵 ∈ ω → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵)))
53	52	imp 122	. 2 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 → suc 𝐴 ⊆ suc 𝐵))
54		nnon 4390	. . 3 ⊢ (𝐴 ∈ ω → 𝐴 ∈ On)
55		onsucsssucr 4292	. . 3 ⊢ ((𝐴 ∈ On ∧ Ord 𝐵) → (suc 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵))
56	54, 25, 55	syl2an 283	. 2 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵))
57	53, 56	impbid 127	1 ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴 ⊆ 𝐵 ↔ suc 𝐴 ⊆ suc 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ↔ wb 103   ∨ wo 662   ∧ w3a 922   = wceq 1287   ∈ wcel 1436  Vcvv 2614   ⊆ wss 2986  ∅c0 3272  Ord word 4156  Oncon0 4157  suc csuc 4159  ωcom 4371

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-10 1439  ax-11 1440  ax-i12 1441  ax-bndl 1442  ax-4 1443  ax-13 1447  ax-14 1448  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067  ax-sep 3925  ax-nul 3933  ax-pow 3977  ax-pr 4003  ax-un 4227  ax-iinf 4369

This theorem depends on definitions:  df-bi 115  df-3an 924  df-tru 1290  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-nfc 2214  df-ne 2252  df-ral 2360  df-rex 2361  df-v 2616  df-dif 2988  df-un 2990  df-in 2992  df-ss 2999  df-nul 3273  df-pw 3411  df-sn 3431  df-pr 3432  df-uni 3631  df-int 3666  df-tr 3905  df-iord 4160  df-on 4162  df-suc 4165  df-iom 4372

This theorem is referenced by:  nnaword  6203