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 ) → ( 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )

Proof

Step Hyp Ref Expression
1 breq1 ( ( abs ‘ 𝑀 ) = 𝑀 → ( ( abs ‘ 𝑀 ) ∥ 𝑁𝑀𝑁 ) )
2 1 bicomd ( ( abs ‘ 𝑀 ) = 𝑀 → ( 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
3 2 a1i ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( abs ‘ 𝑀 ) = 𝑀 → ( 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) ) )
4 negdvdsb ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 𝑀𝑁 ↔ - 𝑀𝑁 ) )
5 breq1 ( ( abs ‘ 𝑀 ) = - 𝑀 → ( ( abs ‘ 𝑀 ) ∥ 𝑁 ↔ - 𝑀𝑁 ) )
6 5 bicomd ( ( abs ‘ 𝑀 ) = - 𝑀 → ( - 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
7 4 6 sylan9bb ( ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) ∧ ( abs ‘ 𝑀 ) = - 𝑀 ) → ( 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )
8 7 ex ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( ( abs ‘ 𝑀 ) = - 𝑀 → ( 𝑀𝑁 ↔ ( 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 ) → ( 𝑀𝑁 ↔ ( abs ‘ 𝑀 ) ∥ 𝑁 ) )