4 color theorem equivalent to cubic planar bridgeless are 3 edge colorable

Suppose we want to 4-color the vertices of a planar graph.

We may assume it's simple, because loops are forbidden and multiple edges don't affect coloring.

We may assume it's a maximal planar graph (that is, a planar triangulation), because adding more edges to triangulate the graph only makes the problem harder.

All planar triangulations on at least 4 vertices are 3-connected.

Instead of coloring the vertices of this graph, we can color the faces of its dual. The dual is another planar graph. It is cubic (because we started with a triangulation) and it is 3-connected (because it's the dual of a simple planar 3-connected graph) so in particular it is bridgeless.

So we have reduced the problem to coloring the faces of a planar cubic bridgeless graph, which is the kind of graph that Tait's theorem applies to.