Metamath Proof Explorer


Theorem divcan3

Description: A cancellation law for division. (Contributed by NM, 3-Feb-2004) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion divcan3 ABB0BAB=A

Proof

Step Hyp Ref Expression
1 eqid BA=BA
2 simp2 ABB0B
3 simp1 ABB0A
4 2 3 mulcld ABB0BA
5 3simpc ABB0BB0
6 divmul BAABB0BAB=ABA=BA
7 4 3 5 6 syl3anc ABB0BAB=ABA=BA
8 1 7 mpbiri ABB0BAB=A