Metamath Proof Explorer


Theorem dmdcan2d

Description: Cancellation law for division and multiplication. (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses div1d.1
|- ( ph -> A e. CC )
divcld.2
|- ( ph -> B e. CC )
divmuld.3
|- ( ph -> C e. CC )
divmuld.4
|- ( ph -> B =/= 0 )
divdiv23d.5
|- ( ph -> C =/= 0 )
Assertion dmdcan2d
|- ( ph -> ( ( A / B ) x. ( B / C ) ) = ( A / C ) )

Proof

Step Hyp Ref Expression
1 div1d.1
 |-  ( ph -> A e. CC )
2 divcld.2
 |-  ( ph -> B e. CC )
3 divmuld.3
 |-  ( ph -> C e. CC )
4 divmuld.4
 |-  ( ph -> B =/= 0 )
5 divdiv23d.5
 |-  ( ph -> C =/= 0 )
6 1 2 4 divcld
 |-  ( ph -> ( A / B ) e. CC )
7 2 3 5 divcld
 |-  ( ph -> ( B / C ) e. CC )
8 6 7 mulcomd
 |-  ( ph -> ( ( A / B ) x. ( B / C ) ) = ( ( B / C ) x. ( A / B ) ) )
9 1 2 3 4 5 dmdcand
 |-  ( ph -> ( ( B / C ) x. ( A / B ) ) = ( A / C ) )
10 8 9 eqtrd
 |-  ( ph -> ( ( A / B ) x. ( B / C ) ) = ( A / C ) )