Editing Invariant (mathematics) (section)

===MU puzzle===
The [[MU puzzle]]<ref>{{Citation | last1 = Hofstadter | first1 = Douglas R. | title = Gödel, Escher, Bach: An Eternal Golden Braid | publisher = Basic Books | year = 1999 | orig-year = 1979 | isbn = 0-465-02656-7 | url-access = registration | url = https://archive.org/details/gdelescherbachet00hofs }}
Here: Chapter I.</ref> is a good example of a logical problem where determining an invariant is of use for an [[impossibility proof]]. The puzzle asks one to start with the word MI and transform it into the word MU, using in each step one of the following transformation rules:

# If a string ends with an I, a U may be appended (''x''I → ''x''IU)
# The string after the M may be completely duplicated (M''x'' → M''xx'')
# Any three consecutive I's (III) may be replaced with a single U (''x''III''y'' → ''x''U''y'')
# Any two consecutive U's may be removed (''x''UU''y'' → ''xy'')

An example derivation (with superscripts indicating the applied rules) is
:MI →<sup>2</sup> MII →<sup>2</sup> MIIII →<sup>3</sup> MUI →<sup>2</sup> MUIUI →<sup>1</sup> MUIUIU →<sup>2</sup> MUIUIUUIUIU →<sup>4</sup> MUIUIIUIU → ...

In light of this, one might wonder whether it is possible to convert MI into MU, using only these four transformation rules. One could spend many hours applying these transformation rules to strings. However, it might be quicker to find a [[Predicate (mathematical logic)|property]] that is invariant to all rules (that is, not changed by any of them), and that demonstrates that getting to MU is impossible. By looking at the puzzle from a logical standpoint, one might realize that the only way to get rid of any I's is to have three consecutive I's in the string. This makes the following invariant interesting to consider:

:''The number of I's in the string is not a multiple of 3''.

This is an invariant to the problem, if for each of the transformation rules the following holds: if the invariant held before applying the rule, it will also hold after applying it. Looking at the net effect of applying the rules on the number of I's and U's, one can see this actually is the case for all rules:

:{| class=wikitable
|-
! Rule !! #I's !! #U's !! Effect on invariant
|-
| style="text-align: center;" | 1 || style="text-align: right;" | +0 || style="text-align: right;" | +1 || Number of I's is unchanged. If the invariant held, it still does.
|-
| style="text-align: center;" |  2 || style="text-align: right;" | ×2 || style="text-align: right;" | ×2 || If ''n'' is not a multiple of 3, then 2×''n'' is not either. The invariant still holds.
|-
| style="text-align: center;" |  3 || style="text-align: right;" | −3 || style="text-align: right;" | +1 || If ''n'' is not a multiple of 3, ''n''−3 is not either. The invariant still holds.
|-
| style="text-align: center;" |  4 || style="text-align: right;" | +0 || style="text-align: right;" | −2 || Number of I's is unchanged. If the invariant held, it still does.
|}

The table above shows clearly that the invariant holds for each of the possible transformation rules, which means that whichever rule one picks, at whatever state, if the number of I's was not a multiple of three before applying the rule, then it  will not be afterwards either.

Given that there is a single I in the starting string MI, and one is not a multiple of three, one can then conclude that it is impossible to go from MI to MU (as the number of I's will never be a multiple of three).