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.