Theorem nnn1suc 43425

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 7428	. . . . . 6 ⊢ (𝑥 = 𝑧 → (𝑥 + 1) = (𝑧 + 1))
23	22	adantl 487	. . . . 5 ⊢ ((𝑧 ∈ ℕ ∧ 𝑥 = 𝑧) → (𝑥 + 1) = (𝑧 + 1))
24	21, 23	rspcedeqvd 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 12357	. 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 1569   ∈ wcel 2144   ≠ wne 2955  ∃wrex 3086  (class class class)co 7421  1c1 11205   + caddc 11207  ℕcn 12339

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pr 5390  ax-un 7752  ax-1cn 11262

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  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 3739  df-csb 3847  df-ss 3896  df-pss 4014  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-pw 4558  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-iun 4952  df-br 5103  df-opab 5167  df-mpt 5186  df-tr 5212  df-id 5545  df-eprel 5550  df-po 5558  df-so 5559  df-fr 5603  df-we 5605  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-pred 6304  df-ord 6365  df-on 6366  df-lim 6367  df-suc 6368  df-iota 6494  df-fun 6541  df-fn 6542  df-f 6543  df-f1 6544  df-fo 6545  df-f1o 6546  df-fv 6547  df-ov 7424  df-om 7879  df-2nd 8003  df-frecs 8303  df-wrecs 8334  df-recs 8383  df-rdg 8422  df-nn 12340

This theorem is used by: (None)