(you should probably read MU puzzle if you havn't already)
There is no
mu puzzle solution - or in the language of MUI
formal system,
mu is not a
theorem of the
axiom mi.
Simply, to reach
mu you must generate a multiple of three number of
i's
The only mechanisms to create and eliminate
i's are doubling (the entire string follow
m,
mx->mxx) and converting three
i's to a
u; Neither by doubling a non-multiple of three (such as 1, in the axiom's case of
mi) nor eliminating groups of threes can we produce a multiple of three.
*shrugs*