**Proof ** The idea is to turn *c*'s
into
*a*'s by **flyping**. A **flype **moves a crossing across
a part of a diagram at the expense of turning the part over:

Notice that if we introduce two crossings using R2 then move one over with a flype we get:

Suppose

Repeat c times to get:

Similarly if *c < 0* * . *If *b
*is
odd *K *will have been rotated through pi about an horizontal
axis.
Use this procedure to eliminate c_{n} , c_{n-1} , ... ,c_{2}
in turn. Then finally:

Notice that a double flype leaves R undisturbed:

We can be more efficient in unwrapping a daisy chain. For example:

After n - 2 double flypes we have:

We now use **flips** to eliminate the unshaded crossings. Here is
a **flip**:

Think of this as turning R over to change the sign of a crossing
between
middle and outer string at the expense of adding a crossing between
middle
and other outer string.

Combine a flip with R2 to get 3 new crossings all of the same sign
as
follows:

Apply R2 flips:

etc.

If n odd get:

which is isotopic to

and if n even get:

which is isotopic to:

**Remark **It follows that the classes of knots defined by
Daisy chains, 4-plats and 2-bridge knots are identical.

See the first fourteen prime knots added together. Notice that they are all 2-bridge knots.

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