4 colors theorem (attempt number 2)

1.  There is no need to use different colors if there is no contact or a common border between to areas(states). So in a group of 4 areas I will need 4 colors only if each 4 areas (states) are in contact with each other. If not, I will need 3 or 2 or 1 color. And  in a group of n areas I need n colors only if each areas are in contact with each other. If not I will need less than n colors.
2. I define a border as a thread lying on the ground for the practical way but in the mathematical way  as a line of point that belong to two  or more areas at the same time . 
3. When the border between two states is only one point (pratical way : a tip) then there is no need to use two different colors for these two states. So we  only care for border having more than one mathemetical point. If the border is only one point we considere for the demonstration that the two areas are only  one area  (the same color can be used).
4. If I can prove that I  can't build  a group of 5 areas, each one  having four borders with the others then I have proved that only four color are needed to differentiate any group of  5 areas.
5. Once a chart with several areas(states) has been built. Then we can change the shape of each area  without changing the fact that there is a border betweeen two states. We could even reduce the length of the border (mathematically the number of points) we just need to stop before we get a tip (a set of one point for the border). The border has  only to keep its'  important charateristic for the demonstration: either it is a  tip (singelton) or a line (not a singleton). In this condition I will not care about the shape of each areas. In this condition if it is ok with one shape then it will be ok for any shape.
6. An other element of the demonstration is to start from an existing plan with n areas and to try to build a copy of this plan   by adding each copies of the areas one by one. We could do it in any order (n areas then n! ways of making the copy). 
7. When building these 5 areas one by one, we can   either build them alone (case a in point 8) or build more  than 5 areas (case b in point 9). But what is important is the condition that link these 5 areas.
8. case a: We try to build these 5 areas  one by one, only by  trying each time to add externally a  new area in contact with all the existing one.  We can see that after having  made 4 of them  then  one is completely enclosed and can not be reached.  If we add a fifth one (just do it with a pen) we can put it in contact with only 3 areas. So only 4 colors would be needed  for any configuration of 5 areas only.
9. case b : we could  try to build n>5 areas and  having a link condition at least  with 5 of them. Such a situation could be build by using the method in case a and add later the supplement inside the shape made by these 4 areas and respecting the primary conditions (each one in contact with each other)  . Would  it change anything to the construction made in case a ? No  because one of the four area would still be unreachable from the outside.  Normally I don't need to know what happen with the supplement areas because I could use the same methode (case b ) with any part of the  n areas  until I reach case a.
10. Now I might be able to prove that a plan with at least one group of 5 areas having common  border with each other doesn't exist: If such a plan  would exist then I could  make a copy of this plan by adding one area after the other. If I begin building   with  the 5 areas of the unique group  then at some point of the construction I would encounter the case a or the case b. So I know I would not be able to make the copy. In this case such a plan doesn't exist and only four colors are enough. 

I think it's a little bit better than the previous version. But is it still a demonstration? I am waiting for some comments .  I am afraid that If some body had understood he would say it