Metamath Proof Explorer


Theorem divcan2zi

Description: A cancellation law for division. (Contributed by NM, 10-Aug-1999)

Ref Expression
Hypotheses divclz.1
|- A e. CC
divclz.2
|- B e. CC
Assertion divcan2zi
|- ( B =/= 0 -> ( B x. ( A / B ) ) = A )

Proof

Step Hyp Ref Expression
1 divclz.1
 |-  A e. CC
2 divclz.2
 |-  B e. CC
3 divcan2
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( B x. ( A / B ) ) = A )
4 1 2 3 mp3an12
 |-  ( B =/= 0 -> ( B x. ( A / B ) ) = A )