Metamath Proof Explorer


Theorem divcan1d

Description: A cancellation 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 )
divcld.3
|- ( ph -> B =/= 0 )
Assertion divcan1d
|- ( ph -> ( ( A / B ) x. B ) = A )

Proof

Step Hyp Ref Expression
1 div1d.1
 |-  ( ph -> A e. CC )
2 divcld.2
 |-  ( ph -> B e. CC )
3 divcld.3
 |-  ( ph -> B =/= 0 )
4 divcan1
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( ( A / B ) x. B ) = A )
5 1 2 3 4 syl3anc
 |-  ( ph -> ( ( A / B ) x. B ) = A )