assumptions - Is my expression too complicated for FullSimplify or am I doing something wrong?

I have a messily defined function $v(h, w)$ with $h, w \in \mathbb{R}$ and with a removable singularity at $h=1/2$, and I am trying to prove some of its properties using Mathematica. In particular I want to prove that $v(h,w) \in \mathbb{R}$, as opposed to possibly having nonzero imaginary part, and I want to prove that $v(h,w) = v(1-h, -w)$. I am defining the function in Mathematica as follows:

a[h_] = h / (2*h - 1)
q[h_, w_] = Sqrt[2*h - 1] * Sqrt[2*w]
v[h_, w_] = Exp[-w]*(2/Sqrt[Pi])*q[h,w]*Exp[(q[h,w]*a[h])^2] / ( Erfi[q[h,w]*a[h]] + Erfi[q[h,w]*(1 - a[h])] )

I apologize for its messiness. My initial hope was that Mathematica would tell me that this is just a well known special function with known properties which I could cite. My next hope was that I could at least use applications of FullSimplify with Assumptions to prove the properties. But I have not been successful at either approach.

To make my question more concrete, I am wondering why Mathematica does not simplify the following expression to True:

FullSimplify[ v[h, w] == v[1 - h, -w], Element[{w,h}, Reals] && h != 1/2]

One possibility is that I am wrong about the math. I've tried to rule this out by using Mathematica to prove special cases like the following, where I've also tried other values of $h$ besides $1/5$.

FullSimplify[ v[1/5, w] / v[1 - 1/5, -w] ]
FullSimplify[ v[1/5, w] - v[1 - 1/5, -w] ]
FullSimplify[ v[1/5, w] == v[1 - 1/5, -w] ]

Another possibility is that it is too hard for Mathematica. But I think the most likely explanation is that I'm using Mathematica wrong because I'm a newbie.

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Update:
Inspired by @rcollyer's answer I took out a pencil and paper and figured out how to write this function as the reciprocal of a series that satisfies the properties by construction:

fterm[h_, w_, k_] = (((-4*w)^k) / (2*k+1)!!) * (h / (2*h-1)) * Exp[w]*(h^2 / (2*h-1))^k
f[h_, w_] = Sum[fterm[h, w, k] + fterm[1-h, -w, k], {k, 0, Infinity}]

Mathematica is still unable to directly verify that $f(h,w) = f(1-h,-w)$ on the appropriate domain, but $1/f(h,w)$ compares True to the original function $v(h,w)$. As a side note, I had hoped that in the process of verifying these properties I would find a formula that gets rid of the computational difficulty associated with the removable singularity at $h=1/2$ for free, but this did not happen.