Associative Nonidentity Einstein Summation

From Schaum's Outline to Tensor Calculus Chapter 1, Example 1.8: $a_{\large{ij}}x_{\large{i}}y_{\large{j}} \neq a_{\large{ij}}y_{\large{i}}x_{\large{j}}$.

I want to tumble to why Einstein Summation shatters the equality without expanding the sums. My two different arguments look too prolix. Is there a faster way?

Because $x_iy_j = y_jx_i,$ I only need to examine the $a_{ij}$'s. Say I sum $j$ first. I reason by fixing the subscripts $i = 1$ and $j = 4$ so I'm examining $a_{\large{14}}$.
On the LHS, $j$ corresponds to $y$ so $y$ gets summed first. So $a_{\large{14}}$ corresponds uniquely to $a_{\large{14}}x_1y_4$.
On the RHS, $j$ corresponds to $x$ so $x$ gets summed first. So $a_{\large{14}}$ corresponds uniquely to $a_{\large{14}}x_4y_1$. But $x_1y_4 \neq x_4y_1$ necessarily.

Another argument? Say I sum $j$ first as before. Say I've already gotten $a_{\large{14}}x_{\large{4}}y_{\large{1}}$ from the RHS.
I want to regenerate $...x_4y_1$ from the LHS. I already supposed summing $j$ first W.L.O.G..
Since $j$ corresponds to $y$ on the LHS, I sum $y$ first. I get $a_{\large{i1}}x_{\large{i}}y_{\large{4}}$. Sum $x$ after to get $a_{\large{41}}x_{\large{i}}y_{\large{4}}$. But $a_{\large{14}} \neq a_{\large{41}}$ necessarily.