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Metamath Proof Explorer

Theorem absdvdsb

Description: An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011)

Ref Expression
Assertion absdvdsb ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )

Proof

Step Hyp Ref Expression
1 breq1 ⊢ ( ( abs ‘ 𝑀 ) = 𝑀 → ( ( abs ‘ 𝑀 ) ∥ 𝑁 ↔ 𝑀 ∥ 𝑁 ) )
2 1 bicomd ⊢ ( ( abs ‘ 𝑀 ) = 𝑀 → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
3 2 a1i ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( abs ‘ 𝑀 ) = 𝑀 → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) ) )
4 negdvdsb ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ 𝑁 ↔ - 𝑀 ∥ 𝑁 ) )
5 breq1 ⊢ ( ( abs ‘ 𝑀 ) = - 𝑀 → ( ( abs ‘ 𝑀 ) ∥ 𝑁 ↔ - 𝑀 ∥ 𝑁 ) )
6 5 bicomd ⊢ ( ( abs ‘ 𝑀 ) = - 𝑀 → ( - 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
7 4 6 sylan9bb ⊢ ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( abs ‘ 𝑀 ) = - 𝑀 ) → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
8 7 ex ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( abs ‘ 𝑀 ) = - 𝑀 → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) ) )
9 zre ⊢ ( 𝑀 ∈ ℤ → 𝑀 ∈ ℝ )
10 9 absord ⊢ ( 𝑀 ∈ ℤ → ( ( abs ‘ 𝑀 ) = 𝑀 ∨ ( abs ‘ 𝑀 ) = - 𝑀 ) )
11 10 adantr ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( abs ‘ 𝑀 ) = 𝑀 ∨ ( abs ‘ 𝑀 ) = - 𝑀 ) )
12 3 8 11 mpjaod ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ 𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )