Reducing system of equations involving Erf, Error Function I have a system of equations: 
$$1/2 + {\rm Erf}(x) - {\rm Erf}(\frac{x+y}{2})=0$$
$$-1/2 + {\rm Erf}(y) - {\rm Erf}(\frac{x+y}{2})=0,$$
Where $x \le y$ and ${\rm Erf}$ is the Error Function.
By ansatz I know that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$, but I can not manage to prove that this is the only solution, even though I strongly suspect it. My question therefore is how to prove uniquness. 
One re-arrangement I have tried results in:
${\rm Erf}(\frac{x+y}{2}) = \frac{{\rm Erf}(x)+{\rm Erf}(y)}{2}$, which I am pretty sure is equivalent to $y=-x$ (apparent by implicit plot), but again I can not prove this. Proving that $y=-x$ for all solutions would imply the desired uniqueness. 
Any advice on manipulation tactics for the error function would be much appreciated..! 
Tagged as probability theory because of the relation to the Normal distribution CDF.
 A: This can be solved as a problem in inequalities.
The desired implication
$$
{\rm Erf}\Bigl(\frac12(x+y)\Bigr) = \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr)
\Longleftrightarrow x + y = 0
$$
is not quite correct, because the hypothesis holds also for $x=y$;
but these are the only possibilities, which is enough to yield
the desired result.  Because the error function is odd, it is enough to prove:
Proposition. If $x$ and $y$ are real numbers such that $x+y > 0$ then
$$
{\rm Erf}\Bigl(\frac12(x+y)\Bigr) \geq \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr),
$$
with equality if and only if $x=y$.
Proof: Let $a = (x+y)/2$ and $b = (x-y)/2$.  Then $a>0$ and $x,y = a \pm b$,
and we claim
$$
\bigl({\rm Erf}(a+b) + {\rm Erf}(a-b)\bigr) - 2\phantom.{\rm Erf}(a) \leq 0,
$$
with equality iff $b=0$.  By symmetry we may assume $b\geq 0$.
Equality clearly holds for $b=0$.  We claim that the left-hand side (LHS)
is a decreasing function of $b$ for $b \geq 0$, which will prove that
it is negative for all $b>0$.  Indeed derivative of the LHS is
$$
\frac2{\sqrt{\pi}} \left( e^{-(a+b)^2} - e^{-(a-b)^2} \right)
= \frac2{\sqrt{\pi}} e^{-(a^2+b^2)} (e^{-2ab} - e^{+2ab}) < 0
$$
for $b>0$, QED.
(This uses the normalization of ${\rm Erf}(t)$ that has derivative
$(2/\sqrt\pi) \phantom. e^{-t^2}$.  There are other normalizations, but the proof
works in the same way.)
A: Here is another proof, arguably more elementary in that it only uses basic properties of $e^{-t^2}$.
Notice that the two original equations reduce to:
$$
\int_x^{(x+y)/2} e^{-t^2}dt = \frac{\sqrt{\pi}}{4}
, \hspace{0.3in}
\int_{(x+y)/2}^y e^{-t^2}dt = \frac{\sqrt{\pi}}{4},
$$
It is clear from this that we must have $y\ne x$. 
With a variable substitution $a=(x+y)/2, b=(x-y)/2$, we also have:
$$
\int_{a-b}^{a} e^{-t^2} dt - \int_{a}^{a+b} e^{-t^2}dt =0.
$$
The following proposition then completes the proof.
Proposition.
For $f(t)$ integrable, positive, symmetric at $t=0$, and increasing for $t<0$, and $b \ne 0$,
$$
\int_{a-b}^{a} f(t)dt - \int_{a}^{a+b} f(t)dt =0 
\hspace{0.2in} \Leftrightarrow \hspace{0.2in}
a=0.
$$
Proof.
The implication $\Leftarrow$ is straight forward. 
To see the implication $\Rightarrow$, notice that a symmetry manipulation gives us (steps omitted):
$$ 
\int_{a-b}^{a} f(t)dt - \int_{a}^{a+b} f(t)dt =
\int_{-a}^{a} f(t)dt - \int_{b-a}^{b+a} f(t)dt.
$$
For $a=0$ this clearly equals $0$. 
For $a>0$, the increasing property and symmetry give us that:
$$
\int_{-a}^{a} f(t)dt > \int_{b-a}^{b+a} f(t)dt.
$$
for all $b \ne 0$. 
An analogous argument holds for $a<0$. QED.
I have not checked wether integrability and positivity are strictly necessary conditions on $f$ for the proposition to hold, but $f(t)=e^{-t^2}$ does have these properties, and generalizations I'm interested in relate to probability distributions, so these assumptions are fine for my purposes. 
