Metamath Proof Explorer


Theorem divcan2i

Description: A cancellation law for division. (Contributed by NM, 9-Feb-1995)

Ref Expression
Hypotheses divclz.1
|- A e. CC
divclz.2
|- B e. CC
divcl.3
|- B =/= 0
Assertion divcan2i
|- ( B x. ( A / B ) ) = A

Proof

Step Hyp Ref Expression
1 divclz.1
 |-  A e. CC
2 divclz.2
 |-  B e. CC
3 divcl.3
 |-  B =/= 0
4 1 2 divcan2zi
 |-  ( B =/= 0 -> ( B x. ( A / B ) ) = A )
5 3 4 ax-mp
 |-  ( B x. ( A / B ) ) = A