This proof is based upon the following premise:
It is known that the 4CT is true if very planar triangulation is 4 vertex colorable.
Every triangulation; ie T-graph has a cubic planar graph as a dual.
Let G be a T-graph and C be the cubic dual of G
Conjecture I: If C is 3 edge colorable, then C is 4 face colorable; and G is 4 vertex colorable.
Conjecture II: If C is the dual of G, then G is the dual of C.
Fact: If C has a Hamiltonian Circuit, then C is 3 edge colorable.
Hypothesis I: If C does not have a HC, then its dual is not a triangulation.
Conjecture: III. If Hypothesis I is true, then G cannot have a HC-less dual.
Conjecture IV. If Conjectures I, II and III are all true, then the 4CT is true.
DISCUSSION:
Conjecture I. If C is 4 face colorable, then G is 4 vertex colorable since
the faces of C map one-to-one with the vertices of C. It is easy to show that
if the edges of C are 3-colorable, then the faces of C are 4-colorable. Therefore,
Conjecture I is true.
Conjecture II. Let G_v, G_e and G_f be the number of vertices, edges and faces
in G. Let C_v, C_e and C_f be the number of vertices, edges and faces in C. If
the dual/dual conjecture is true; then G_f = C_v and C_f = G_v.
By Euler's Formula G_f = G_e - G_v + 2. In T-graph G_e = 3*(G_v) - 6. Therefore
1) G_f = 3*(G_v)-6 - G_v +2 = 2*(G_v) - 4
2) C_v = G_f = 2*(G_v) - 4
In a cubic there are 1.5 edges per vertex. Therefore,
3) C_e = 1.5*C_v. Again, using Euler's formula;
4) C_f = C_e - C_v + 2 = 1.5*C_v - C_v + 2
= 0.5*C_v + 2 = G_v
If we know G_v we can calculate G_f, C_v and C_f. If C_f = G_v, then ??
From
4) C_f = 0.5*(C_v) + 2
2) C_v = G_f and
2) G_f = 2*(G-v) - 4 , we get
5) C_f = 0.5*(G_f) + 2
C_f = 0.5*(2*(G_v-4) + 2
+2