Metamath Proof Explorer


Theorem zdivmul

Description: Property of divisibility: if D divides A then it divides B x. A . (Contributed by NM, 3-Oct-2008)

Ref Expression
Assertion zdivmul
|- ( ( ( D e. NN /\ A e. ZZ /\ B e. ZZ ) /\ ( A / D ) e. ZZ ) -> ( ( B x. A ) / D ) e. ZZ )

Proof

Step Hyp Ref Expression
1 zcn
 |-  ( B e. ZZ -> B e. CC )
2 1 3ad2ant2
 |-  ( ( A e. ZZ /\ B e. ZZ /\ D e. NN ) -> B e. CC )
3 zcn
 |-  ( A e. ZZ -> A e. CC )
4 3 3ad2ant1
 |-  ( ( A e. ZZ /\ B e. ZZ /\ D e. NN ) -> A e. CC )
5 nncn
 |-  ( D e. NN -> D e. CC )
6 nnne0
 |-  ( D e. NN -> D =/= 0 )
7 5 6 jca
 |-  ( D e. NN -> ( D e. CC /\ D =/= 0 ) )
8 7 3ad2ant3
 |-  ( ( A e. ZZ /\ B e. ZZ /\ D e. NN ) -> ( D e. CC /\ D =/= 0 ) )
9 divass
 |-  ( ( B e. CC /\ A e. CC /\ ( D e. CC /\ D =/= 0 ) ) -> ( ( B x. A ) / D ) = ( B x. ( A / D ) ) )
10 2 4 8 9 syl3anc
 |-  ( ( A e. ZZ /\ B e. ZZ /\ D e. NN ) -> ( ( B x. A ) / D ) = ( B x. ( A / D ) ) )
11 10 3comr
 |-  ( ( D e. NN /\ A e. ZZ /\ B e. ZZ ) -> ( ( B x. A ) / D ) = ( B x. ( A / D ) ) )
12 11 adantr
 |-  ( ( ( D e. NN /\ A e. ZZ /\ B e. ZZ ) /\ ( A / D ) e. ZZ ) -> ( ( B x. A ) / D ) = ( B x. ( A / D ) ) )
13 zmulcl
 |-  ( ( B e. ZZ /\ ( A / D ) e. ZZ ) -> ( B x. ( A / D ) ) e. ZZ )
14 13 3ad2antl3
 |-  ( ( ( D e. NN /\ A e. ZZ /\ B e. ZZ ) /\ ( A / D ) e. ZZ ) -> ( B x. ( A / D ) ) e. ZZ )
15 12 14 eqeltrd
 |-  ( ( ( D e. NN /\ A e. ZZ /\ B e. ZZ ) /\ ( A / D ) e. ZZ ) -> ( ( B x. A ) / D ) e. ZZ )