To follow on [How to prove Tait's theorem about planar cubic bridgeless graph being 3-edge-colorable?
The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.
I disagree with the solution given (As stated in my comment). The provided links does not proove the equivalence. It shows 1) from 4 color-theorem, how to build a 3-edge coloring for bridgeless cubic graph 2) from a 3-edge-coloring, how to build a 4 face coloring for the same graph
The theorem by Tait is much more powerful. If I can 3-edge color any cubic bridgeless planar graph, then I can 4-color ANY planar graph (not just cubic bridgeless planar).
Any idea how to prove the equivalence. I cannot find the original paper from Tait. Lots of reference but never the actual proof. The implication 4CT $\Rightarrow$ 3-edge-coloring for bridgeless planar cubic graph is easy. The other implication is the one missing : $$ \{ \forall G, \text{ cubic, planar, bridgeless}, \exists \text{ a 3-edge coloring}\}$$ $$\Rightarrow$$ $$\{\forall G, \text{ planar,} \ \exists \text{a 4-vertex-coloring}\}$$