Calculating impulse response of system Given is the illustrated circuit diagram of a linear, time-invariant, time-discrete system:



How do I show that the total system has the impulse response $h[n] = aδ[n] + bδ [n - 1] + cδ [n - 2]$ and determine the constants a,b and c?

with the following impulse responses of the subsystems:

$h_1[n] = δ[n] + (1/2)δ[n − 1]$

$h_2[n] = δ[n] − (1/2)δ[n − 1]$

$h_3[n] = −δ[n]$

$h4[n] = − (1/2)^n σ[n]$

So what I have done:

$h_{3||4}=−δ[n]−(1/2)^n σ[n]$

and than:

$h_{3||4+2}=\sum_{k=-\infty}^{\infty}(δ[n-k] − (1/2)δ[n-k − 1])(−δ[n]−(1/2)^n σ[n])$

$h_{3||4+2}=−δ[n]−(1/2)^n σ[n]-\frac{1}{2}(−δ[n-1]−(1/2)^{n-1} σ[n-1])=−δ[n]-\frac{1}{2}^nδ[n]+\frac{1}{2}δ[n-1]$

$h_{3||4+2+1}=(\frac{1}{2})^nδ[n]-δ[n-1]$

But the solution is $a=1$ and $b=-1$. How is $a=1$?
Working in z-domain makes the problem easy 

$$ H_{1}(z) = 1 + \frac{1}{2}z^{-1} = \frac{2z+1}{2z}$$
$$ H_{2}(z) = 1 - \frac{1}{2}z^{-1} = \frac{2z-1}{2z}$$
$$ H_{3}(z) = -1$$
$$ H_{4}(z) = \frac{-z}{z - \frac{1}{2}} = \frac{-2z}{2z-1}$$

$$ H_{3}(z) + H_{4}(z) = \frac{-4z + 1}{2z + 1} $$ 
$$ H_{2}(H_{3} + H_{4}) = (\frac{2z-1}{2z})(\frac{1-4z}{2z-1}) = \frac{1-4z}{2z}$$

$$ H(z) = H_{1} || H_{2}(H_{3} + H_{4}) = \frac{2z+1}{2z} + \frac{1-4z}{2z} =\frac{1-z}{z} $$
$$ H(z) =  z^{-1} - 1 $$
$$ \implies h[n] = -\delta[n] + \delta[n-1] $$
Hence, $a = -1, b = 1, c = 0$