Twisted convolution - simplifying multiple integral via substitution I need help with understanding a proof from a paper. Following is the description:

Let
$$r_{1}(t) = \int \int h_{1}\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} s\left(t - \tau\right) d\tau d\nu$$
$$r(t) = \int \int h_{2}\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} r_{1}\left(t - \tau\right) d\tau d\nu$$
Substituting $r_{1}(t)$ in $r_{2}(t)$  we obtain after some algebraic
manipulations,
$$r(t) = \int \int h\left(\tau, \nu\right) e^{j 2\pi \nu (t - \tau)} s\left(t - \tau\right) d\tau d\nu$$
where, $f(\tau, \nu)$ is given by
$$h(\tau, \nu) = \int \int h_{2}\left(\tau', \nu' \right) h_{1}\left(\tau - \tau', \nu-\nu' \right) e^{j 2\pi \nu' \left(\tau - \tau' \right)} d\tau' d\nu'$$

I am unable to use appropriate algebraic manipulations to arrive at the final expression for $r(t)$. Can anyone help me? The first step is arrived at by substituting $r_{1}(t-\tau)$ in $r_{2}(t)$. But I am unable to proceed further. (Incorporated corrections from Tanya Choudhary's answer)
$$r(t) = \int \int h_2(\tau,\nu)e^{j 2\pi\nu(t-\tau)} \int \int h_1(\tau_1,\nu_1)e^{j 2\pi\nu(t-\tau-\tau_1)}s(t-\tau-\tau_1)d\tau_1 d\nu_1 d\tau d\nu$$
Substituting $\tau = \tau', \tau_1 = \tau - \tau', \nu = \nu', \nu_1 = \nu - \nu'$,
$$r(t) = \int \int h_2(\tau',\nu')e^{j 2\pi\nu'(t-\tau')} \int \int h_1(\tau - \tau',\nu - \nu')e^{j 2\pi\nu(t-\tau)}s(t-\tau)d\tau' d\nu' d\tau d\nu$$
Since these are two different integral equations, during substitution you cannot use the same variables for both the integrals.
On substituting $r_1(t-\tau)$ to $r(t)$ we get,
$$r(t) = \int \int h_2(\tau,\nu)e^{j 2\pi\nu(t-\tau)} \int \int h_1(\tau_1,\nu_1)e^{j 2\pi\nu(t-\tau-\tau_1)}s(t-\tau-\tau_1)d\tau_1 d\nu_1 d\tau d\nu$$
Now substitute $\tau = \tau', \tau_1 = \tau - \tau', \nu = \nu', \nu_1 = \nu - \nu'$