Theorem nnn1suc 43597

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 3021	. . . 4 ⊢ (𝑦 = 1 → (𝑦 ≠ 1 ↔ 1 ≠ 1))
2		eqeq2 2775	. . . . 5 ⊢ (𝑦 = 1 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 1))
3	2	rexbidv 3190	. . . 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 3021	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑦 ≠ 1 ↔ 𝑧 ≠ 1))
6		eqeq2 2775	. . . . 5 ⊢ (𝑦 = 𝑧 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝑧))
7	6	rexbidv 3190	. . . 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 3021	. . . 4 ⊢ (𝑦 = (𝑧 + 1) → (𝑦 ≠ 1 ↔ (𝑧 + 1) ≠ 1))
10		eqeq2 2775	. . . . 5 ⊢ (𝑦 = (𝑧 + 1) → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = (𝑧 + 1)))
11	10	rexbidv 3190	. . . 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 3021	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ≠ 1 ↔ 𝐴 ≠ 1))
14		eqeq2 2775	. . . . 5 ⊢ (𝑦 = 𝐴 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝐴))
15	14	rexbidv 3190	. . . 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 2960	. . . 4 ⊢ (1 ≠ 1 ↔ ¬ 1 = 1)
18		eqid 2763	. . . . 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 7458	. . . . . 6 ⊢ (𝑥 = 𝑧 → (𝑥 + 1) = (𝑧 + 1))
23	22	adantl 487	. . . . 5 ⊢ ((𝑧 ∈ ℕ ∧ 𝑥 = 𝑧) → (𝑥 + 1) = (𝑧 + 1))
24	21, 23	rspcedeqvd 3587	. . . 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 12393	. 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 2147   ≠ wne 2959  ∃wrex 3090  (class class class)co 7451  1c1 11240   + caddc 11242  ℕcn 12375

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-nul 5266  ax-pr 5398  ax-un 7783  ax-1cn 11297

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 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-reu 3370  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-pss 4017  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-iun 4958  df-br 5109  df-opab 5174  df-mpt 5193  df-tr 5219  df-id 5553  df-eprel 5558  df-po 5566  df-so 5567  df-fr 5611  df-we 5613  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-rn 5669  df-res 5670  df-ima 5671  df-pred 6330  df-ord 6391  df-on 6392  df-lim 6393  df-suc 6394  df-iota 6520  df-fun 6568  df-fn 6569  df-f 6570  df-f1 6571  df-fo 6572  df-f1o 6573  df-fv 6574  df-ov 7454  df-om 7912  df-2nd 8036  df-frecs 8337  df-wrecs 8368  df-recs 8417  df-rdg 8456  df-nn 12376

This theorem is used by: (None)