Metamath Proof Explorer


Theorem divcan1

Description: A cancellation law for division. (Contributed by NM, 5-Jun-2004) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion divcan1 ABB0ABB=A

Proof

Step Hyp Ref Expression
1 divcl ABB0AB
2 simp2 ABB0B
3 1 2 mulcomd ABB0ABB=BAB
4 divcan2 ABB0BAB=A
5 3 4 eqtrd ABB0ABB=A