Metamath Proof Explorer


Theorem divdiv3d

Description: Division into a fraction. (Contributed by Glauco Siliprandi, 24-Dec-2020)

Ref Expression
Hypotheses divdiv3d.1
|- ( ph -> A e. CC )
divdiv3d.2
|- ( ph -> B e. CC )
divdiv3d.3
|- ( ph -> C e. CC )
divdiv3d.4
|- ( ph -> B =/= 0 )
divdiv3d.5
|- ( ph -> C =/= 0 )
Assertion divdiv3d
|- ( ph -> ( ( A / B ) / C ) = ( A / ( C x. B ) ) )

Proof

Step Hyp Ref Expression
1 divdiv3d.1
 |-  ( ph -> A e. CC )
2 divdiv3d.2
 |-  ( ph -> B e. CC )
3 divdiv3d.3
 |-  ( ph -> C e. CC )
4 divdiv3d.4
 |-  ( ph -> B =/= 0 )
5 divdiv3d.5
 |-  ( ph -> C =/= 0 )
6 1 2 3 4 5 divdiv1d
 |-  ( ph -> ( ( A / B ) / C ) = ( A / ( B x. C ) ) )
7 2 3 mulcomd
 |-  ( ph -> ( B x. C ) = ( C x. B ) )
8 7 oveq2d
 |-  ( ph -> ( A / ( B x. C ) ) = ( A / ( C x. B ) ) )
9 6 8 eqtrd
 |-  ( ph -> ( ( A / B ) / C ) = ( A / ( C x. B ) ) )