Solution to MIU puzzle:
Count the number of I's in a theorem, it seems it is never 0,
in fact it is never a multiple of 3. This may mean that
MIII can never evolve into MU.
(1) # of I's is 1 to start with != multiple of 3.
(2) Rules 1 and 4 do not affect I Count.
(3) Rule 3 diminishes I count by 3 exactly
= only create a multiple of 3 if already a multiple of 3.
Rule 2 doubles I count
if 2n is multiple of 3
then n must be multiple.
Therefore I count cannot be a multiple of 3.
Therefore MIII --> MU is not possible.
Therefore cannot get an I count of 0.
And hence MU is not in system.