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Metamath Proof Explorer

Theorem nn1m1nn

Description: Every positive integer is one or a successor. (Contributed by Mario Carneiro, 16-May-2014)

Ref Expression
Assertion nn1m1nn ( 𝐴 ∈ ℕ → ( 𝐴 = 1 ∨ ( 𝐴 − 1 ) ∈ ℕ ) )

Proof

Step Hyp Ref Expression
1 orc ⊢ ( 𝑥 = 1 → ( 𝑥 = 1 ∨ ( 𝑥 − 1 ) ∈ ℕ ) )
2 1cnd ⊢ ( 𝑥 = 1 → 1 ∈ ℂ )
3 1 2 2thd ⊢ ( 𝑥 = 1 → ( ( 𝑥 = 1 ∨ ( 𝑥 − 1 ) ∈ ℕ ) ↔ 1 ∈ ℂ ) )
4 eqeq1 ⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 1 ↔ 𝑦 = 1 ) )
5 oveq1 ⊢ ( 𝑥 = 𝑦 → ( 𝑥 − 1 ) = ( 𝑦 − 1 ) )
6 5 eleq1d ⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 − 1 ) ∈ ℕ ↔ ( 𝑦 − 1 ) ∈ ℕ ) )
7 4 6 orbi12d ⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 = 1 ∨ ( 𝑥 − 1 ) ∈ ℕ ) ↔ ( 𝑦 = 1 ∨ ( 𝑦 − 1 ) ∈ ℕ ) ) )
8 eqeq1 ⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( 𝑥 = 1 ↔ ( 𝑦 + 1 ) = 1 ) )
9 oveq1 ⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( 𝑥 − 1 ) = ( ( 𝑦 + 1 ) − 1 ) )
10 9 eleq1d ⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( ( 𝑥 − 1 ) ∈ ℕ ↔ ( ( 𝑦 + 1 ) − 1 ) ∈ ℕ ) )
11 8 10 orbi12d ⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( ( 𝑥 = 1 ∨ ( 𝑥 − 1 ) ∈ ℕ ) ↔ ( ( 𝑦 + 1 ) = 1 ∨ ( ( 𝑦 + 1 ) − 1 ) ∈ ℕ ) ) )
12 eqeq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 = 1 ↔ 𝐴 = 1 ) )
13 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 − 1 ) = ( 𝐴 − 1 ) )
14 13 eleq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 − 1 ) ∈ ℕ ↔ ( 𝐴 − 1 ) ∈ ℕ ) )
15 12 14 orbi12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 = 1 ∨ ( 𝑥 − 1 ) ∈ ℕ ) ↔ ( 𝐴 = 1 ∨ ( 𝐴 − 1 ) ∈ ℕ ) ) )
16 ax-1cn ⊢ 1 ∈ ℂ
17 nncn ⊢ ( 𝑦 ∈ ℕ → 𝑦 ∈ ℂ )
18 pncan ⊢ ( ( 𝑦 ∈ ℂ ∧ 1 ∈ ℂ ) → ( ( 𝑦 + 1 ) − 1 ) = 𝑦 )
19 17 16 18 sylancl ⊢ ( 𝑦 ∈ ℕ → ( ( 𝑦 + 1 ) − 1 ) = 𝑦 )
20 id ⊢ ( 𝑦 ∈ ℕ → 𝑦 ∈ ℕ )
21 19 20 eqeltrd ⊢ ( 𝑦 ∈ ℕ → ( ( 𝑦 + 1 ) − 1 ) ∈ ℕ )
22 21 olcd ⊢ ( 𝑦 ∈ ℕ → ( ( 𝑦 + 1 ) = 1 ∨ ( ( 𝑦 + 1 ) − 1 ) ∈ ℕ ) )
23 22 a1d ⊢ ( 𝑦 ∈ ℕ → ( ( 𝑦 = 1 ∨ ( 𝑦 − 1 ) ∈ ℕ ) → ( ( 𝑦 + 1 ) = 1 ∨ ( ( 𝑦 + 1 ) − 1 ) ∈ ℕ ) ) )
24 3 7 11 15 16 23 nnind ⊢ ( 𝐴 ∈ ℕ → ( 𝐴 = 1 ∨ ( 𝐴 − 1 ) ∈ ℕ ) )