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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 )