Contradiction while using the convolution sum for a non-LTI system In a recent quiz, we were given the following problem:


  The cascaded LTI systems $\mathcal{T}_1$ and $\mathcal{T}_2$ respectively have impulse responses $h_1[n]=\delta[n+3]$ and $h_2[n]=\delta[5−n]$. What is the output when the input is $x[n]=n$, i.e., find $y[n]=\mathcal{T}_2\{\mathcal{T}_1\{x[n]\}\}$.


I am primarily confused about the "LTI-ness" of the system and at a contradiction; given the second stage to be LTI (assuming that for a system to be LTI as a whole, all its stages/subsystems must also be LTI). Thus, we can say that a sub-system with  $h \lbrack n \rbrack = \delta \lbrack 5 - n\rbrack$, i.e. $y \lbrack n \rbrack = x \lbrack 5 - n\rbrack$ should be LTI, which already seems to be false. To prove that, I considered the following convolution sum: $$\tilde{y} \lbrack n \rbrack = \displaystyle{\sum_{k = -\infty}^{\infty} x \lbrack k \rbrack \delta \lbrack 5 - n + k \rbrack} = x \lbrack n-5 \rbrack \ne x \lbrack 5-n \rbrack = y \lbrack n \rbrack$$

And this is a contradiction since, after the convolution, the output doesn't match the original output with which we started.

Also, I realized that interestingly, the convolution always gives a result that corresponds to an LTI system (as in the case above too: $y[n]=x[n−5]$ is LTI whereas $y[n] = x[5-n]$ is not). 

Also, since $\delta[n]$ is even, any non-LTI system's impulse response, for example, $\delta[1−n]$, will equal $\delta[n−1]$ which corresponds the impulse response of an LTI system. This explains why I am getting LTI characteristics after the convolution. This is interesting too since in either way the impulse response implies delaying the signal by 1 (in the current example). 

And so did the impulse response $\delta[5-n]$ stated in the quiz question just qualitatively imply a delay of 5, and is, technically, not the exact description of the underlying system?

So, in summary, I have the following two doubts:


Is the convolution sum only true for the output relation of an LTI system? If so, then can it be used to prove non-LTI-ness of a system in the same manner as above?
Is the quiz question incorrect?


Any help will be greatly appreciated!
You've some troubles about the LTI property and the impulse response. First of all let me give you the answer to your quiz question.

Given two LTI systems $\mathcal{T}_1$ and $\mathcal{T}_2$ with impulse responses :

$$h_1[n] = \delta[n+3]$$
$$h_2[n] = \delta[5-n] = \delta[n-5]$$

you can compute the output of the cascade $$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} $$ for the input signal $x[n] = n$ as :

$$ y_1[n] = \mathcal{T}_1 \{ x[n] \} \implies y_1[n] = n \star \delta[n+3] = (n+3) $$

and 

$$ y_2[n] = \mathcal{T}_2 \{ y_1[n] \} \implies y_2[n] = (n+3) \star \delta[n-5] = (n-2) $$

so your output is $$y[n] = n-2$$

Note that $h_2[n] = \delta[5-n]$ and $h_2[n] = \delta[n-5]$ are the same systems, and yield the same output. 

Note again that both systems are declared to be LTI and are given their respective impulse responses $h_1[n]$ and $h_2[n]$ , therefore their cascade will also be LTI and furthermore the output does not depend on the ordering of the systems; i.e.

$$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} = \mathcal{T}_1 \{ \mathcal{T}_2 \{ x[n] \} \}  $$

Now coming to your own questions:

1- You cannot use convolution sum to deduce LTI property. You cannot form a convolution sum if the system is not already LTI.

2- The question seems ok:

Note: your claim that $ h[n] = \delta[5-n] $ , signifies a non-LTI system is wrong...