Metamath Proof Explorer


Theorem divcan6

Description: Cancellation of inverted fractions. (Contributed by NM, 28-Dec-2007)

Ref Expression
Assertion divcan6 AA0BB0ABBA=1

Proof

Step Hyp Ref Expression
1 recdiv AA0BB01AB=BA
2 1 oveq2d AA0BB0AB1AB=ABBA
3 divcl ABB0AB
4 3 3expb ABB0AB
5 4 adantlr AA0BB0AB
6 divne0 AA0BB0AB0
7 recid ABAB0AB1AB=1
8 5 6 7 syl2anc AA0BB0AB1AB=1
9 2 8 eqtr3d AA0BB0ABBA=1