Metamath Proof Explorer


Theorem divcan1d

Description: A cancellation law for division. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 φ A
divcld.2 φ B
divcld.3 φ B 0
Assertion divcan1d φ A B B = A

Proof

Step Hyp Ref Expression
1 div1d.1 φ A
2 divcld.2 φ B
3 divcld.3 φ B 0
4 divcan1 A B B 0 A B B = A
5 1 2 3 4 syl3anc φ A B B = A