Metamath Proof Explorer


Theorem divassd

Description: An associative law for 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 )
divassd.4
|- ( ph -> C =/= 0 )
Assertion divassd
|- ( ph -> ( ( A x. B ) / C ) = ( A x. ( B / 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 divassd.4
 |-  ( ph -> C =/= 0 )
5 divass
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A x. B ) / C ) = ( A x. ( B / C ) ) )
6 1 2 3 4 5 syl112anc
 |-  ( ph -> ( ( A x. B ) / C ) = ( A x. ( B / C ) ) )