Here, we should not be able to commute each l through m without picking up some unknown commutators. However, this outputs l_{a} m_{b} l_{c}, suggesting that the algorithm assumes that, since l commutes with itself, this expression is symmetric in a,c. This sadly isn't sufficient, so we get the wrong answer.
Note that if alternatively l does not commute with itself, this issue is dodged:
If I understand correctly, consider this (v2.2.0):
Here, we should not be able to commute each
l
throughm
without picking up some unknown commutators. However, this outputsl_{a} m_{b} l_{c}
, suggesting that the algorithm assumes that, sincel
commutes with itself, this expression is symmetric ina,c
. This sadly isn't sufficient, so we get the wrong answer.Note that if alternatively
l
does not commute with itself, this issue is dodged:(Prints
l_{c} m_{b} l_{a}
.)