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 ) )