(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.


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