How to prove Tait's theorem about planar cubic bridgeless graph being 3-edge-colorable? How can be proved, that
The  four-color  theorem  is  equivalent  to  the  claim  that  every planar cubic bridgeless graph is 3-edge-colorable.
I can't figure out or find any prove of this theorem.
Thanks.
 A: The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.
A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.
Since the four color problem has been already proved, also the three edge coloring is true.
A: The four colors can be represented by bit-vectors of length 2.
We can define $3$ transformations, $L, R, B$ on these bit-vectors, where
$L$ inverts the left-bit, $R$ inverts the right-bit, and $B$ inverts both bits.
Each of these is bijective with no fixed point.
Note that composing all three of these gives the identity,
and composing any two of them gives the third.
Now consider a maximal planar graph. If the vertices can be $4$-colored
then each edge determines one of these transformations and the edges
of a triangle must have all $3$ transformations in order to compose to the
identity transformation.
Conversely, if the edges can be $3$-colored such that the $3$ edges of
each triangle have different colors, then $L,R,B$ can be used for the colors,
and the transformations can be used to extend a color on a starting vertex
to the whole graph.
A: I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:
$$(a, b) \equiv (c,d) \equiv 1$$
$$(a, c) \equiv (b,d) \equiv 2$$
$$(a, d) \equiv (b,c) \equiv 3$$
In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors  around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.
