Theorem nnn1suc 43147

Description: A positive integer that is not 1 is a successor of some other positive integer. (Contributed by Steven Nguyen, 19-Aug-2023.)

Assertion

Ref	Expression
nnn1suc	⊢ ((𝐴 ∈ ℕ ∧ 𝐴 ≠ 1) → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴)

Distinct variable group:   𝑥,𝐴

Proof of Theorem nnn1suc

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

Step	Hyp	Ref	Expression
1		neeq1 3017	. . . 4 ⊢ (𝑦 = 1 → (𝑦 ≠ 1 ↔ 1 ≠ 1))
2		eqeq2 2772	. . . . 5 ⊢ (𝑦 = 1 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 1))
3	2	rexbidv 3186	. . . 4 ⊢ (𝑦 = 1 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 1))
4	1, 3	imbi12d 347	. . 3 ⊢ (𝑦 = 1 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (1 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)))
5		neeq1 3017	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑦 ≠ 1 ↔ 𝑧 ≠ 1))
6		eqeq2 2772	. . . . 5 ⊢ (𝑦 = 𝑧 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝑧))
7	6	rexbidv 3186	. . . 4 ⊢ (𝑦 = 𝑧 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧))
8	5, 7	imbi12d 347	. . 3 ⊢ (𝑦 = 𝑧 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (𝑧 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧)))
9		neeq1 3017	. . . 4 ⊢ (𝑦 = (𝑧 + 1) → (𝑦 ≠ 1 ↔ (𝑧 + 1) ≠ 1))
10		eqeq2 2772	. . . . 5 ⊢ (𝑦 = (𝑧 + 1) → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = (𝑧 + 1)))
11	10	rexbidv 3186	. . . 4 ⊢ (𝑦 = (𝑧 + 1) → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1)))
12	9, 11	imbi12d 347	. . 3 ⊢ (𝑦 = (𝑧 + 1) → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ ((𝑧 + 1) ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))))
13		neeq1 3017	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ≠ 1 ↔ 𝐴 ≠ 1))
14		eqeq2 2772	. . . . 5 ⊢ (𝑦 = 𝐴 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝐴))
15	14	rexbidv 3186	. . . 4 ⊢ (𝑦 = 𝐴 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴))
16	13, 15	imbi12d 347	. . 3 ⊢ (𝑦 = 𝐴 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (𝐴 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴)))
17		df-ne 2956	. . . 4 ⊢ (1 ≠ 1 ↔ ¬ 1 = 1)
18		eqid 2760	. . . . 5 ⊢ 1 = 1
19	18	pm2.24i 151	. . . 4 ⊢ (¬ 1 = 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)
20	17, 19	sylbi 220	. . 3 ⊢ (1 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)
21		id 23	. . . . 5 ⊢ (𝑧 ∈ ℕ → 𝑧 ∈ ℕ)
22		oveq1 7420	. . . . . 6 ⊢ (𝑥 = 𝑧 → (𝑥 + 1) = (𝑧 + 1))
23	22	adantl 487	. . . . 5 ⊢ ((𝑧 ∈ ℕ ∧ 𝑥 = 𝑧) → (𝑥 + 1) = (𝑧 + 1))
24	21, 23	rspcedeq1vd 3583	. . . 4 ⊢ (𝑧 ∈ ℕ → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))
25	24	2a1d 27	. . 3 ⊢ (𝑧 ∈ ℕ → ((𝑧 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧) → ((𝑧 + 1) ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))))
26	4, 8, 12, 16, 20, 25	nnind 12275	. 2 ⊢ (𝐴 ∈ ℕ → (𝐴 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴))
27	26	imp 412	1 ⊢ ((𝐴 ∈ ℕ ∧ 𝐴 ≠ 1) → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ≠ wne 2955  ∃wrex 3086  (class class class)co 7413  1c1 11125   + caddc 11127  ℕcn 12257

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732  ax-sep 5251  ax-nul 5263  ax-pr 5398  ax-un 7736  ax-1cn 11182

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-reu 3366  df-rab 3413  df-v 3452  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-pss 3919  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-iun 4953  df-br 5104  df-opab 5168  df-mpt 5187  df-tr 5213  df-id 5550  df-eprel 5555  df-po 5563  df-so 5564  df-fr 5608  df-we 5610  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-pred 6299  df-ord 6360  df-on 6361  df-lim 6362  df-suc 6363  df-iota 6489  df-fun 6535  df-fn 6536  df-f 6537  df-f1 6538  df-fo 6539  df-f1o 6540  df-fv 6541  df-ov 7416  df-om 7863  df-2nd 7987  df-frecs 8280  df-wrecs 8311  df-recs 8360  df-rdg 8399  df-nn 12258

This theorem is used by: (None)