Theorem nnn1suc 43066

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 3022	. . . 4 ⊢ (𝑦 = 1 → (𝑦 ≠ 1 ↔ 1 ≠ 1))
2		eqeq2 2777	. . . . 5 ⊢ (𝑦 = 1 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 1))
3	2	rexbidv 3191	. . . 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 3022	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑦 ≠ 1 ↔ 𝑧 ≠ 1))
6		eqeq2 2777	. . . . 5 ⊢ (𝑦 = 𝑧 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝑧))
7	6	rexbidv 3191	. . . 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 3022	. . . 4 ⊢ (𝑦 = (𝑧 + 1) → (𝑦 ≠ 1 ↔ (𝑧 + 1) ≠ 1))
10		eqeq2 2777	. . . . 5 ⊢ (𝑦 = (𝑧 + 1) → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = (𝑧 + 1)))
11	10	rexbidv 3191	. . . 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 3022	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ≠ 1 ↔ 𝐴 ≠ 1))
14		eqeq2 2777	. . . . 5 ⊢ (𝑦 = 𝐴 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝐴))
15	14	rexbidv 3191	. . . 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 2961	. . . 4 ⊢ (1 ≠ 1 ↔ ¬ 1 = 1)
18		eqid 2765	. . . . 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 7423	. . . . . 6 ⊢ (𝑥 = 𝑧 → (𝑥 + 1) = (𝑧 + 1))
23	22	adantl 487	. . . . 5 ⊢ ((𝑧 ∈ ℕ ∧ 𝑥 = 𝑧) → (𝑥 + 1) = (𝑧 + 1))
24	21, 23	rspcedeq1vd 3590	. . . 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 12262	. 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 2146   ≠ wne 2960  ∃wrex 3091  (class class class)co 7416  1c1 11112   + caddc 11114  ℕcn 12244

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-nul 5271  ax-pr 5406  ax-un 7738  ax-1cn 11169

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 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ne 2961  df-ral 3082  df-rex 3092  df-reu 3372  df-rab 3419  df-v 3459  df-sbc 3747  df-csb 3855  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-pss 3926  df-nul 4287  df-if 4490  df-pw 4566  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-iun 4960  df-br 5112  df-opab 5176  df-mpt 5195  df-tr 5221  df-id 5558  df-eprel 5563  df-po 5571  df-so 5572  df-fr 5616  df-we 5618  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-pred 6306  df-ord 6367  df-on 6368  df-lim 6369  df-suc 6370  df-iota 6496  df-fun 6542  df-fn 6543  df-f 6544  df-f1 6545  df-fo 6546  df-f1o 6547  df-fv 6548  df-ov 7419  df-om 7865  df-2nd 7989  df-frecs 8280  df-wrecs 8311  df-recs 8360  df-rdg 8399  df-nn 12245

This theorem is used by: (None)