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 e. ZZ /\ N e. ZZ ) -> ( 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 e. ZZ /\ N e. ZZ ) -> ( ( abs ` M ) = M -> ( M || N <-> ( abs ` M ) || N ) ) )
4 negdvdsb 
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( M || N <-> -u M || N ) )
5 breq1 
 |-  ( ( abs ` M ) = -u M -> ( ( abs ` M ) || N <-> -u M || N ) )
6 5 bicomd 
 |-  ( ( abs ` M ) = -u M -> ( -u M || N <-> ( abs ` M ) || N ) )
7 4 6 sylan9bb 
 |-  ( ( ( M e. ZZ /\ N e. ZZ ) /\ ( abs ` M ) = -u M ) -> ( M || N <-> ( abs ` M ) || N ) )
8 7 ex 
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( ( abs ` M ) = -u M -> ( M || N <-> ( abs ` M ) || N ) ) )
9 zre 
 |-  ( M e. ZZ -> M e. RR )
10 9 absord 
 |-  ( M e. ZZ -> ( ( abs ` M ) = M \/ ( abs ` M ) = -u M ) )
11 10 adantr 
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( ( abs ` M ) = M \/ ( abs ` M ) = -u M ) )
12 3 8 11 mpjaod 
 |-  ( ( M e. ZZ /\ N e. ZZ ) -> ( M || N <-> ( abs ` M ) || N ) )