Metamath Proof Explorer


Theorem ddcan

Description: Cancellation in a double division. (Contributed by Mario Carneiro, 26-Apr-2015)

Ref Expression
Assertion ddcan AA0BB0AAB=B

Proof

Step Hyp Ref Expression
1 simpll AA0BB0A
2 simprl AA0BB0B
3 simprr AA0BB0B0
4 divcan1 ABB0ABB=A
5 1 2 3 4 syl3anc AA0BB0ABB=A
6 divcl ABB0AB
7 1 2 3 6 syl3anc AA0BB0AB
8 divne0 AA0BB0AB0
9 divmul ABABAB0AAB=BABB=A
10 1 2 7 8 9 syl112anc AA0BB0AAB=BABB=A
11 5 10 mpbird AA0BB0AAB=B