A metric cone $C$ is a nonempty metric space (whose distance is denoted $d$) together with a map $\cdot\colon \mathbf{R}\times C \mapsto C$ satisfying these axioms:


* $a\cdot(b\cdot x) = (ab)\cdot x$ for all reals $a$ and $b$, and all $x$ in $C$,
* $d(a\cdot x;a\cdot y) = \vert a\vert d(x;y)$ for all real $a$ and all $x$ and $y$ in $C$,
* $d(a\cdot x;b\cdot x) = \vert a-b\vert d(x;0)$ for all reals $a$ and $b$, and all $x$ in $C$, where $0$ denote $0\cdot x$ for any $x \in C$ (this definition is independent of the choice of $x$),
* $d((a+a')\cdot x;(b+b')\cdot y) \leq d(a\cdot x;b\cdot y) + d((a'\cdot x;b'\cdot y)$ for all reals $a$, $b$, $a'$, and $b'$, and all $x$ and $y$ in $C$.


Those structures arise in my work about metric vector spaces. In some situations, I manage to prove the three first axioms, but not the fourth. So my question is:



> 
> Is the fourth axiom independent from the three other ones?
> 
> 
> 


I have looked for a counterexample, and for a demonstration of the last axiom from the previous, but I failed in both attempts.



Independence of the axiomatics of metric cones
Yes, it is independent.


Consider function $f\colon \mathbb R^2\to \mathbb R\_\ge$ defened the following way:
$$f(x,y)=f(y,x)$$
and if $|y|\le |x|$ then
$$f(x,y)=|x|+\min\{|y|,|y-\tfrac12{\cdot} x|\}.$$
Note that one can define a metric on the union of two coordinate lines $C=\{(x,y)\in \mathbb R^2\mid x{\cdot}y=0\}$ the following way:
$$d((x,0),(y,0))=d((0,x),(0,y))=|x-y|$$
and
$$d((x,0),(0,y))=f(x,y).$$


It is straightforward to check that your first axioms hold for $d$.
Take $x=(1,0)$, $y=(0,1)$, $a=1$, $b=\tfrac12$, $a'=\tfrac12$ and $b'=1$.
Then 
$$d((a+a')\cdot x;(b+b')\cdot y)=\tfrac32+\tfrac34>2.$$


On the other hand 
$$d(a\cdot x;b\cdot y) + d((a'\cdot x;b'\cdot y)=2.$$


