TITLE: about addition and multiplications of real number
QUESTION [18 upvotes]: We all know that for every two real number $x,y$, the operation of addition $x+y$ satisfies the following conditions:
\begin{gather}
x+y=y+x;\\
x+0=x;\\
(x+y)+z=x+(y+z);\\
x+(-x)=0.
\end{gather}
Now, take a function $f\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying the following four conditions:

$\forall x,y\in \Bbb R, f(x,y)=f(y,x);$
$\forall x\in \Bbb R, f(x,0)=x;$
$\forall x,y,z\in \Bbb R, f(f(x,y),z)=f(x,f(y,z));$
$\forall x\in \Bbb R, f(x,-x)=0.$

Can we conclude that $f(x,y)=x+y$ for all $x,y\in \Bbb R$?
edit:  Thanks to Jared, the answer is "NO". 
think  a function $g\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying the following four conditions:
a). $\forall x,y\in \Bbb R, g(x,y)=g(y,x);$
b). $\forall x\in \Bbb R, g(x,1)=x;$
c). $\forall x,y,z\in \Bbb R, g(g(x,y),z)=g(x,g(y,z));$
d). $\forall x\in \Bbb R,x\ne0,  g(x,\frac1x)=1.$
Can we conclude that $g(x,y)=xy$ for all $x,y\in \Bbb R$?
again, two functions $f,g\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying 1,2,3,4, a),b),c),d),and
e). $\forall x,y,z\in \Bbb R, g(x,f(y,z))=f(g(x,y),g(x,z))$
Can we conclude that $f(x,y)=x+y,g(x,y)=xy$ for all $x,y\in \Bbb R$? It seems that the answer is NO, $f(x,y)=\sqrt[3]{x^3+y^3},g(x,y)=xy$ works

REPLY [21 votes]: Counterexample:
$$f(x,y)=\sqrt[3]{x^3+y^3}$$