Theorem nnsucsssuc 6388

Description: Membership is inherited by successors. The reverse direction holds for all ordinals, as seen at onsucsssucr 4425, but the forward direction, for all ordinals, implies excluded middle as seen as onsucsssucexmid 4442. (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 3120	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵))
2		suceq 4324	. . . . . . 7 ⊢ (𝑥 = 𝐴 → suc 𝑥 = suc 𝐴)
3	2	sseq1d 3126	. . . . . 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 3120	. . . . . 6 ⊢ (𝑥 = ∅ → (𝑥 ⊆ 𝐵 ↔ ∅ ⊆ 𝐵))
7		suceq 4324	. . . . . . 7 ⊢ (𝑥 = ∅ → suc 𝑥 = suc ∅)
8	7	sseq1d 3126	. . . . . 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 3120	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝑥 ⊆ 𝐵 ↔ 𝑦 ⊆ 𝐵))
11		suceq 4324	. . . . . . 7 ⊢ (𝑥 = 𝑦 → suc 𝑥 = suc 𝑦)
12	11	sseq1d 3126	. . . . . 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 3120	. . . . . 6 ⊢ (𝑥 = suc 𝑦 → (𝑥 ⊆ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
15		suceq 4324	. . . . . . 7 ⊢ (𝑥 = suc 𝑦 → suc 𝑥 = suc suc 𝑦)
16	15	sseq1d 3126	. . . . . 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 4510	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → suc 𝐵 ≠ ∅)
19	18	neneqd 2329	. . . . . . . 8 ⊢ (𝐵 ∈ ω → ¬ suc 𝐵 = ∅)
20		peano2 4509	. . . . . . . . . 10 ⊢ (𝐵 ∈ ω → suc 𝐵 ∈ ω)
21		0elnn 4532	. . . . . . . . . 10 ⊢ (suc 𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
22	20, 21	syl 14	. . . . . . . . 9 ⊢ (𝐵 ∈ ω → (suc 𝐵 = ∅ ∨ ∅ ∈ suc 𝐵))
23	22	ord 713	. . . . . . . 8 ⊢ (𝐵 ∈ ω → (¬ suc 𝐵 = ∅ → ∅ ∈ suc 𝐵))
24	19, 23	mpd 13	. . . . . . 7 ⊢ (𝐵 ∈ ω → ∅ ∈ suc 𝐵)
25		nnord 4525	. . . . . . . 8 ⊢ (𝐵 ∈ ω → Ord 𝐵)
26		ordsucim 4416	. . . . . . . 8 ⊢ (Ord 𝐵 → Ord suc 𝐵)
27		0ex 4055	. . . . . . . . 9 ⊢ ∅ ∈ V
28		ordelsuc 4421	. . . . . . . . 9 ⊢ ((∅ ∈ V ∧ Ord suc 𝐵) → (∅ ∈ suc 𝐵 ↔ suc ∅ ⊆ suc 𝐵))
29	27, 28	mpan 420	. . . . . . . 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 983	. . . . . . . . . 10 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → suc 𝑦 ⊆ 𝐵)
34		simp1l 1005	. . . . . . . . . . 11 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵) ∧ suc 𝑦 ⊆ 𝐵) → 𝑦 ∈ ω)
35		simp1r 1006	. . . . . . . . . . . 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 4421	. . . . . . . . . . 11 ⊢ ((𝑦 ∈ ω ∧ Ord 𝐵) → (𝑦 ∈ 𝐵 ↔ suc 𝑦 ⊆ 𝐵))
38	34, 36, 37	syl2anc 408	. . . . . . . . . 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 6387	. . . . . . . . . 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 4509	. . . . . . . . . 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 4421	. . . . . . . . 9 ⊢ ((suc 𝑦 ∈ ω ∧ Ord suc 𝐵) → (suc 𝑦 ∈ suc 𝐵 ↔ suc suc 𝑦 ⊆ suc 𝐵))
47	44, 45, 46	syl2anc 408	. . . . . . . 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 1183	. . . . . 6 ⊢ (((𝑦 ∈ ω ∧ 𝐵 ∈ ω) ∧ (𝑦 ⊆ 𝐵 → suc 𝑦 ⊆ suc 𝐵)) → (suc 𝑦 ⊆ 𝐵 → suc suc 𝑦 ⊆ suc 𝐵))
50	49	exp31 361	. . . . 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 4515	. . . 4 ⊢ (𝑥 ∈ ω → (𝐵 ∈ ω → (𝑥 ⊆ 𝐵 → suc 𝑥 ⊆ suc 𝐵)))
52	5, 51	vtoclga 2752	. . 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 4523	. . 3 ⊢ (𝐴 ∈ ω → 𝐴 ∈ On)
55		onsucsssucr 4425	. . 3 ⊢ ((𝐴 ∈ On ∧ Ord 𝐵) → (suc 𝐴 ⊆ suc 𝐵 → 𝐴 ⊆ 𝐵))
56	54, 25, 55	syl2an 287	. 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 697   ∧ w3a 962   = wceq 1331   ∈ wcel 1480  Vcvv 2686   ⊆ wss 3071  ∅c0 3363  Ord word 4284  Oncon0 4285  suc csuc 4287  ωcom 4504

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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-13 1491  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121  ax-sep 4046  ax-nul 4054  ax-pow 4098  ax-pr 4131  ax-un 4355  ax-iinf 4502

This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ne 2309  df-ral 2421  df-rex 2422  df-v 2688  df-dif 3073  df-un 3075  df-in 3077  df-ss 3084  df-nul 3364  df-pw 3512  df-sn 3533  df-pr 3534  df-uni 3737  df-int 3772  df-tr 4027  df-iord 4288  df-on 4290  df-suc 4293  df-iom 4505

This theorem is referenced by:  nnaword  6407  ennnfonelemk  11913  ennnfonelemkh  11925