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Metamath Proof Explorer

Theorem dvdsdivcl

Description: The complement of a divisor of N is also a divisor of N . (Contributed by Mario Carneiro, 2-Jul-2015) (Proof shortened by AV, 9-Aug-2021)

Ref Expression
Assertion dvdsdivcl ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } ) → ( 𝑁 / 𝐴 ) ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } )

Proof

Step Hyp Ref Expression
1 breq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∥ 𝑁 ↔ 𝐴 ∥ 𝑁 ) )
2 1 elrab ⊢ ( 𝐴 ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } ↔ ( 𝐴 ∈ ℕ ∧ 𝐴 ∥ 𝑁 ) )
3 nndivdvds ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ ℕ ) → ( 𝐴 ∥ 𝑁 ↔ ( 𝑁 / 𝐴 ) ∈ ℕ ) )
4 3 biimpd ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ ℕ ) → ( 𝐴 ∥ 𝑁 → ( 𝑁 / 𝐴 ) ∈ ℕ ) )
5 4 expcom ⊢ ( 𝐴 ∈ ℕ → ( 𝑁 ∈ ℕ → ( 𝐴 ∥ 𝑁 → ( 𝑁 / 𝐴 ) ∈ ℕ ) ) )
6 5 com23 ⊢ ( 𝐴 ∈ ℕ → ( 𝐴 ∥ 𝑁 → ( 𝑁 ∈ ℕ → ( 𝑁 / 𝐴 ) ∈ ℕ ) ) )
7 6 imp ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐴 ∥ 𝑁 ) → ( 𝑁 ∈ ℕ → ( 𝑁 / 𝐴 ) ∈ ℕ ) )
8 nnne0 ⊢ ( 𝐴 ∈ ℕ → 𝐴 ≠ 0 )
9 8 anim1ci ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐴 ∥ 𝑁 ) → ( 𝐴 ∥ 𝑁 ∧ 𝐴 ≠ 0 ) )
10 divconjdvds ⊢ ( ( 𝐴 ∥ 𝑁 ∧ 𝐴 ≠ 0 ) → ( 𝑁 / 𝐴 ) ∥ 𝑁 )
11 9 10 syl ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐴 ∥ 𝑁 ) → ( 𝑁 / 𝐴 ) ∥ 𝑁 )
12 7 11 jctird ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐴 ∥ 𝑁 ) → ( 𝑁 ∈ ℕ → ( ( 𝑁 / 𝐴 ) ∈ ℕ ∧ ( 𝑁 / 𝐴 ) ∥ 𝑁 ) ) )
13 2 12 sylbi ⊢ ( 𝐴 ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } → ( 𝑁 ∈ ℕ → ( ( 𝑁 / 𝐴 ) ∈ ℕ ∧ ( 𝑁 / 𝐴 ) ∥ 𝑁 ) ) )
14 13 impcom ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } ) → ( ( 𝑁 / 𝐴 ) ∈ ℕ ∧ ( 𝑁 / 𝐴 ) ∥ 𝑁 ) )
15 breq1 ⊢ ( 𝑥 = ( 𝑁 / 𝐴 ) → ( 𝑥 ∥ 𝑁 ↔ ( 𝑁 / 𝐴 ) ∥ 𝑁 ) )
16 15 elrab ⊢ ( ( 𝑁 / 𝐴 ) ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } ↔ ( ( 𝑁 / 𝐴 ) ∈ ℕ ∧ ( 𝑁 / 𝐴 ) ∥ 𝑁 ) )
17 14 16 sylibr ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } ) → ( 𝑁 / 𝐴 ) ∈ { 𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁 } )