GEB: Is MUUI a MIU-string?

Question

This is a sub-puzzle of the MIU-system, as described by Douglas Hofstadter in his book Gödel, Escher, Bach. (One can also find a description of this system here.)

My question is: is MUUI a MIU-string (assuming, as usual, that the sole axiom is the string MI)?

If yes, I'd like to see a derivation, starting from MI. If no, I'd like to see a proof.

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NB: As a reminder, the MIU-system rules may not be "run backwards". In particular, even though rule IV allows one to drop UU from any MIU-string, it does not allow one to add UU to any MIU string. In other words, rule IV may not be invoked to obtain MUUI from the axiom string MI.

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EDIT

The following is excerpted from p. 260 of the latest American edition of GEB:

SYMBOLS: M, I, U

AXIOM: MI

RULES:
I. If xI is a theorem, so is xIU.
II. If Mx is a theorem, so is Mxx.
III. In any theorem, III can be replaced by U.
IV. UU can be dropped from any theorem.

NB:

In the rules,

1. the word "theorem" is synonym for what I call "MIU-string" earlier in the post;
2. the variable x stands for any MIU-substring.

See here for examples, and for additional details.

Answer 1

It is in fact possible to obtain any string $Mx$ where $x$ consists of $I$s and $U$s and its number of $I$s is not a multiple of $3$ (conversely, it is easy to see you cannot obtain any string not of this form). First, I claim it is possible to make $MI^n$ for any $n$ that is not a multiple of $3$. To do this, let $2^m>n$ be such that $2^m-n$ is a multiple of $3$ (this is possible since $n$ is not a multiple of $3$). Using rule II $m$ times gives $MI^{2^m}$. If $2^m-n$ is a multiple of $6$, it is now easy to reach $MI^n$, since you can delete copies of $I^6$ using rules III and IV. If $2^m-n$ is not a multiple of $6$, note that you can first use rule I to get $MI^{2^m}U$, then use rules III and IV to get $MI^{2^m-3}$. From there you can remove copies of $I^6$ to get $MI^n$.

Now, given any string $Mx$ where $x$ consists of $I$s and $U$s and its number of $I$s is not a multiple of $3$, let $a$ be the number of $I$s in $x$ and let $b$ be the number of $U$s in $x$. Then $n=a+3b$ is not a multiple of $3$, so we can make $MI^n$. Replacing copies of $I^3$ with $U$, we can then obtain $Mx$.

For an explicit derivation in the case of $MUUI$, we start by using rule II four times to get to $MI^{16}$. Then rule I gives $MI^{16}U$, rule III gives $MI^{13}UU$, and rule IV gives $MI^{13}$. Using rule III and rule IV we delete an $I^6$ to get $MI^7$. Finally, we use rule III twice to get $MUUI$.

Answer 2

One possible solution:

$$MI \overset{2}{\rightarrow} MII \overset{2}{\rightarrow} MIIII \overset{2}{\rightarrow} MIIIIIIII \overset{1}{\rightarrow} MIIIIIIIIU \overset{3}{\rightarrow} MIIIIIUU \overset{4}{\rightarrow} MIIIII \overset{2}{\rightarrow} $$

$$MIIIIIIIIII \overset{1}{\rightarrow} MIIIIIIIIIIU \overset{3}{\rightarrow} MIIIIIIIUU \overset{4}{\rightarrow} MIIIIIII \overset{3}{\rightarrow} MUIIII \overset{3}{\rightarrow} MUUI$$