For some $\mathcal{C}_i = [n, k_i, d_i]$ in $\mathbb{F}_2$ where $i\in\{1,2\}$ the $(u | u+v)$ construction is defined as \begin{align*} \mathcal{C} = \{(u\ |\ u+v)\ |\ u\in\mathcal{C}_1,\ v\in\mathcal{C}_2\} \end{align*}
I have to derive the deriving the minimum distance of this code, and the value that all other derivations* give is $d_{\min} = \min(2d_1, d_2)$. However in all these derivations the case where $v=u\neq 0$ is not considered. In this case you would have \begin{align*} ||(u\ |\ u+v)|| = ||u\ |\ 2u|| = ||u\ |\ 0|| \geq d_1 \end{align*} So in the cases where $u$ and $v$ can be equal, $d_{\min} = \min(2d_1, d_2)$ would not hold. Instead it should be $d_{\min} = \min(d_1, d_2)$.
Is there anything wrong with the counterproof I made, or is there a particular reason that they do not consider the case where $v=u\neq 0$?
*Like the derivation on page 28 here, or the derivation from page 4 onward here.
If $u=v$, then $u$ is a word of the code $\mathcal{C}_2$. Hence $(u|0)=(v|0)$ has weight at least $d_2\ge\min\{2d_1,d_2\}$, so the claim still holds.
My experience may be a bit more limited than it should, but I'm under the impression that the $(u|u+v)$ construction is typically used when $d_2>d_1$ and $\mathcal{C}_2\subset \mathcal{C}_1$ — the Reed-Muller codes being a well known case. I recall being a bit surprised to learn that the bound holds more generally.