I believe that your analysis is correct. Or, those match my own thoughts, anyway.
]]>You are correct. You must double the entire string succeeding the 'M' when applying rule 2:
MIUIIUIU -> MIUIIUIUIUIIUIU when applying rule 2
MIIUUI -> MIIUUIIIUUI when applying rule 2
etc.
]]>Edit: I think this may prove it:
Interesting puzzle though. It doesn't look trivial to prove either way.
]]>I assume this is somehow wrong:
MI -> MII -> MIII -> MU
]]>Anyway, this post is about a particular puzzle that is given at the first of the book. I have not finished the book, so I'm not sure whether or not there is a solution; however, I thought everyone might enjoy churning over it.
So, here it is:
1. Begin with the string 'MI'.
2. Now, you can modify your string at any point according to the following rules:
2a. If your string ends in an 'I', you can add a 'U' to the end of it.
2b. If your string is of the form M{therest}, you can add {therest} to the end, creating M{therest}{therest}.
2c. If your string contains 'III', the 'III' can be replaced with a 'U'.
2d. If your string contains 'UU', the 'UU' can be dropped.
3. Use the rules above and change 'MI' into 'MU'.
To make sure that it's clear, here is a possible way to apply the rules:
MI -> MII (rule 2b) -> MIIII (rule 2b) -> MIIIIU (rule 2a) -> MIUU (rule 2c) -> MI (rule 2d)
Can you get 'MU'?
Anyway, I have my own thoughts about the answer to the above question, but I will withhold them temporarily as to not color your own investigations.
The discussion around the puzzle is very interesting as well, since it is given specifically to teach about axioms, theorems, etc., but perhaps I will save that for another thread...
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