Changing order of summation - proof How was the right side of equation obtained from its left side? I could obviously guess immediately that this is true, but mathematics is not about guessing. Are there any intermediate steps between left and right side to make it obvious?

 A: My justification would be as follows.
Given this:
$$\sum_k \left (\sum_l x[l]h_1[n-k-l]\right)h_2[k]$$
Notice what the evaluations of $x$ and $h_2$ each depend on. $x$ depends exclusively on $l$, and $h_2$ depends exclusively on $k$, respectively. Thus, with respect to the other indexer ($k$ for $l$, and vice versa), the evaluation is a constant (i.e. $\frac{\partial x}{\partial k}=\frac{\partial h_2}{\partial l}=0$).
As you know, the following identity (summing times a constant equals the constant times the sum) holds:
$$\sum_a c\cdot w[a]=(c \cdot w_1+c \cdot w_2+c \cdot w_3...)=c \cdot (w_1+w_2+w_3+...)$$
$$=c \cdot \sum_a w[a]$$
We can now apply that same reasoning to the original expression:
$$\sum_k \left (\sum_l x[l]h_1[n-k-l]\right)h_2[k]$$
$$=(x[1] \cdot h_1[n-1-1]+\color{red} {x[2] \cdot h_1[n-1-2]}+\color{blue} {x[3] \cdot h_1[n-1-3]}+...) \cdot h_2[1]+\dots$$
$$(\color{red} {x[1] \cdot h_1[n-2-1]}+x[2] \cdot h_1[n-2-2]+\color{green} {x[3] \cdot h_1[n-2-3]}+...) \cdot h_2[2]+\dots$$
$$(\color{blue} {x[1] \cdot h_1[n-3-1]}+\color{green} {x[2] \cdot h_1[n-3-2]}+x[3] \cdot h_1[n-3-3]+...) \cdot h_2[3]+\dots$$
$$\vdots$$
Notice the symmetry across the diagonal. If you were to order things along the columns and sum, the results would be the same. Thus, you can swap the orders of the sum for the terms which are dependent on only one sum. That's why the $h_1[n-l-k]$ cannot move.
