This question is related to my previous post: [New edge coloring problem in graph theory](https://mathoverflow.net/q/283886/90655). 


*Added:* Let $G$ be a simple graph. Consider the following edge coloring:


1. We are allowed to use repetitive colors on some edges incident to a vertex such that the result does not contain a sequence of length $3$ of one color.
2. The maximum different colors used for coloring the edges incident to a vertex is $s$.


Obviously $s\leq \Delta$. Is the following clime true? Could someone provide a counterexample or a sketch of proof for this clime?



> 
> **Clime (Edited after Chris Godsil Answer):** Edges of every simple graph can be colored due to above coloring with at most $s+1$ color. where $s=\lfloor\frac{\Delta+1}{2}\rfloor$+1.
> 
> 
> 



Edges of every simple graph can be colored with at most $s+1$ color
It might be less confusing if you just claimed/asked if $\lfloor \frac{\Delta+3}2\rfloor$ colors suffice. However it is not correct. The $10$ edges of the complete graph $K\_5$ can not be $3$ colored without creating a monochromatic $3$ edge path (which might be closed, i.e. a triangle.)


Here is a sketch which seems a little long but does the job:


1. Note that $K\_4$ can , as claimed, be colored with $3$ colors. However $2$ colors are not enough.
2. In a possible $3$ edge coloring of $K\_5$ observe that no vertex can touch $3$ (or $4$) edges of the same color because then the edges of the remaining vertices are restricted to the other two colors, and that contradicts 1.
3. There are at least $4$ edges with the same color, say $a.$ Suppose $12,13$ are colored $a$. Then neither of $14,15$ is. Also none of $23,24,25,34,35$ can have color $a.$ But that leaves only $45$ which (so far) could have color $a$ and that is only three edges, not four.


