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Theorem nnn1suc 42557
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.
StepHypRef Expression
1 neeq1 2995 . . . 4 (𝑦 = 1 → (𝑦 ≠ 1 ↔ 1 ≠ 1))
2 eqeq2 2749 . . . . 5 (𝑦 = 1 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 1))
32rexbidv 3161 . . . 4 (𝑦 = 1 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 1))
41, 3imbi12d 344 . . 3 (𝑦 = 1 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (1 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)))
5 neeq1 2995 . . . 4 (𝑦 = 𝑧 → (𝑦 ≠ 1 ↔ 𝑧 ≠ 1))
6 eqeq2 2749 . . . . 5 (𝑦 = 𝑧 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝑧))
76rexbidv 3161 . . . 4 (𝑦 = 𝑧 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧))
85, 7imbi12d 344 . . 3 (𝑦 = 𝑧 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (𝑧 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧)))
9 neeq1 2995 . . . 4 (𝑦 = (𝑧 + 1) → (𝑦 ≠ 1 ↔ (𝑧 + 1) ≠ 1))
10 eqeq2 2749 . . . . 5 (𝑦 = (𝑧 + 1) → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = (𝑧 + 1)))
1110rexbidv 3161 . . . 4 (𝑦 = (𝑧 + 1) → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1)))
129, 11imbi12d 344 . . 3 (𝑦 = (𝑧 + 1) → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ ((𝑧 + 1) ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))))
13 neeq1 2995 . . . 4 (𝑦 = 𝐴 → (𝑦 ≠ 1 ↔ 𝐴 ≠ 1))
14 eqeq2 2749 . . . . 5 (𝑦 = 𝐴 → ((𝑥 + 1) = 𝑦 ↔ (𝑥 + 1) = 𝐴))
1514rexbidv 3161 . . . 4 (𝑦 = 𝐴 → (∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦 ↔ ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴))
1613, 15imbi12d 344 . . 3 (𝑦 = 𝐴 → ((𝑦 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑦) ↔ (𝐴 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴)))
17 df-ne 2934 . . . 4 (1 ≠ 1 ↔ ¬ 1 = 1)
18 eqid 2737 . . . . 5 1 = 1
1918pm2.24i 150 . . . 4 (¬ 1 = 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)
2017, 19sylbi 217 . . 3 (1 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 1)
21 id 22 . . . . 5 (𝑧 ∈ ℕ → 𝑧 ∈ ℕ)
22 oveq1 7367 . . . . . 6 (𝑥 = 𝑧 → (𝑥 + 1) = (𝑧 + 1))
2322adantl 481 . . . . 5 ((𝑧 ∈ ℕ ∧ 𝑥 = 𝑧) → (𝑥 + 1) = (𝑧 + 1))
2421, 23rspcedeq1vd 3584 . . . 4 (𝑧 ∈ ℕ → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))
25242a1d 26 . . 3 (𝑧 ∈ ℕ → ((𝑧 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝑧) → ((𝑧 + 1) ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = (𝑧 + 1))))
264, 8, 12, 16, 20, 25nnind 12167 . 2 (𝐴 ∈ ℕ → (𝐴 ≠ 1 → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴))
2726imp 406 1 ((𝐴 ∈ ℕ ∧ 𝐴 ≠ 1) → ∃𝑥 ∈ ℕ (𝑥 + 1) = 𝐴)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395   = wceq 1542  wcel 2114  wne 2933  wrex 3061  (class class class)co 7360  1c1 11031   + caddc 11033  cn 12149
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709  ax-sep 5242  ax-nul 5252  ax-pr 5378  ax-un 7682  ax-1cn 11088
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3or 1088  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-mo 2540  df-eu 2570  df-clab 2716  df-cleq 2729  df-clel 2812  df-nfc 2886  df-ne 2934  df-ral 3053  df-rex 3062  df-reu 3352  df-rab 3401  df-v 3443  df-sbc 3742  df-csb 3851  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-pss 3922  df-nul 4287  df-if 4481  df-pw 4557  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-iun 4949  df-br 5100  df-opab 5162  df-mpt 5181  df-tr 5207  df-id 5520  df-eprel 5525  df-po 5533  df-so 5534  df-fr 5578  df-we 5580  df-xp 5631  df-rel 5632  df-cnv 5633  df-co 5634  df-dm 5635  df-rn 5636  df-res 5637  df-ima 5638  df-pred 6260  df-ord 6321  df-on 6322  df-lim 6323  df-suc 6324  df-iota 6449  df-fun 6495  df-fn 6496  df-f 6497  df-f1 6498  df-fo 6499  df-f1o 6500  df-fv 6501  df-ov 7363  df-om 7811  df-2nd 7936  df-frecs 8225  df-wrecs 8256  df-recs 8305  df-rdg 8343  df-nn 12150
This theorem is referenced by: (None)
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