Metamath Proof Explorer


Theorem divcan7d

Description: Cancel equal divisors in a division. (Contributed by Mario Carneiro, 27-May-2016)

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 divcan7d
|- ( ph -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )

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 divcan7
 |-  ( ( A e. CC /\ ( B e. CC /\ B =/= 0 ) /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )
7 1 2 4 3 5 6 syl122anc
 |-  ( ph -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )