Metamath Proof Explorer


Theorem divdiv1d

Description: Division into a fraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 φ A
divcld.2 φ B
divmuld.3 φ C
divmuld.4 φ B 0
divdiv23d.5 φ C 0
Assertion divdiv1d φ A B C = A B C

Proof

Step Hyp Ref Expression
1 div1d.1 φ A
2 divcld.2 φ B
3 divmuld.3 φ C
4 divmuld.4 φ B 0
5 divdiv23d.5 φ C 0
6 divdiv1 A B B 0 C C 0 A B C = A B C
7 1 2 4 3 5 6 syl122anc φ A B C = A B C