For moves of type I things are quite simple. Suppose that the following
part of a coloured
knot is given and that the shown arcs are coloured *a* and *b*.
If a move of type I is applied to it then thetwo arcs become a single arc
and it is not clear whether the new single arc should be assigned the colour
*a* or the colour *b*. Fortunately, no choice has to be made
because, by using the defintion of a colouring, it is possible to show
that *a*= *b*. To see that this is so, botice that at the crossing
shown the equation *a* + *b* = 2? *a mod m* must be satisfied.
Regardless of the value of *m*, by subtracting *a* from both
sides of the equation it follows that *b* = *a mod m*. A Reidemeister
move of type I which adds an arc by flipping the string around; namely
which transforms
into is even easier
to handle. In this case, simply colour both new arcs the same colour as
the original arc from which they both originated.

Moves of type II are a bit more complicated. Suppose that part of a
knot has been coloured as follows
It suffices to show that the new knot after the Reidemeister move of type
II has been applied it can be coloured by giving the new arc colour *a
*and keeping the colour of the old arc the same. One difficulty which
might arise is that thepart which was oroiginally coloured
*c *is
now coloured *a *and this migth cause problems in the rest of the
knot. However, it will be shown that
*a *= *c *, so that this
problem does not occuur. Notice
that from the definition of a colouring the equations
*a *+ *b *=2?
*d mod m* and
*c *+ *b *=2? *d mod m*. As in the case
of type I moves, the value *m* is not important because, from the
two equations it follows that
*c *+ *b *= *a *+ *b mod
m *and, hence, *c *= *a mod m *. The other problem which might
arice is that, by eliminating the colour
*b *the number of colours
has been reduced to 1 (recall that a colouring musthave at least two distinct
colours). However, if this is the case then *d *= *a mod m *and,
recalling that
*a *+ *b *=2? *d mod m*, it follows that
*b
= d mod m*. But now it follows that all all fours colours *a *,
*b *, *c *and *d *are the same modulo *m *. Hence the
total numer of colours used is can not be reduced by using only *a *and
*d *. The Reidemeisetr moves of type II in the other direction are
more easily handled, Simply assign the new arc the colour it had before.

Finally, Reidemeister moves of type III must be considered. Consider
part of a knot, shown on the left,
with the colouring given by
*a *, *b *, *c *, *d *,
*e *and *f *. A Reidemeister move of type II will transform this
part of the knot as follows
and, if the arcs shown are assigned the indicated colouring, the question
becomes: "Is it possible to find a value *x* so that the resulting
assignment is a valid colouring?". By considering the crossing at which
the colours *a *, *c *and *x *meet, it is clear that *x
*must be equal to 2 ? *a *- *c *. It must be check that this
value of *x *does not violate the colouring equation at the crossing
where the colours *e *, *b
*and *x *meet. This equation
is:

(*) 2 ? *a *- *c *+ *e *= 2 ? *b*

From the colouring equation at the crossing where the colours *e *,
*d
*and *a *meet, it follows that 2 ? *a *= *d *+ *e
*and by considering the colouring equation at the crossing where the
colours *c *, *d
*and *f *meet, it follows that 2 ? *f
*= *d *+ *c *. By subtracting the second equation from the
first, one gets that *-c *+ *e *= 2 ? *a *- 2 ? *f *.
By substituting this into (*) one gets that 2 ? *a *+ ( 2 ? *a *-
2 ? *f *) = 2 ? *b *and this reduces to 4 ? *a *= 2 ? *f
*+ 2 ? *b *. This equation holds by looking at the crossing where
the colours *a *, *f
*and *b *meet. This shows that this
assignment of *x
*results in a good colouring. The other Reidemeister
move of type III is handled similarly.