Binary addition preserving Hamming weights Let $x,y$ be two $n$-bit strings, with Hamming weights (number of $1$ bits) equal to $w_{1},w_{2}$, respectively. Let $z$ be the binary representation of the sum $x+y$, where we interpret $x$ and $y$ as binary numbers. If the two strings $x$ and $y$ do not share a position where they both have a $1$ (that is, if there is no $1\leq{}i\leq{}n$ such that the $i^{th}$ bits of both $x$ and $y$ are $1$) then it is clear that the Hamming weight of $z$ is exactly $w_{1}+w_{2}$. 
My question is: is the converse also true? Is it true that if the Hamming weight of the binary representation $z$ of the sum $x+y$ is exactly $w_{1}+w_{2}$, then $x$ and $y$ do not share a position where they both have a $1$? 
This looks to be true to me, but how does one go about proving this? I tried arguing about how a carry during addition would mess things up, but this gets messy soon. Perhaps a correct induction hypothesis is what I am missing here? 
 A: Yes, and this holds more generally for the sum of digits in any base $b$.$\DeclareMathOperator{\digsum}{digitsum}$
Let $a_k$ be the $k$th digit of $a$, so $a = \sum a_k b^k$. Then

$\digsum_b(a+c) \leq \digsum_b(a) + \digsum_b(c)$, with equality if and only if $a_k + c_k < b$ in every position $k$ (i.e. the sum has
  no "carries").

Proof: Let $d = a + c$ and $t$ be the number of digits in $d$.
Define $i_{-1} = 0$,
and for $k = 0, \dots, t$ let $i_k$ be 1 if there is a carry from position $k$ to position $k+1$, i.e. if $a_k + c_k + i_{k-1} \geq b$,
and 0 otherwise.
Then
\begin{align}
\digsum(d) &= \sum_{k=0}^t d_k\\
&= \sum_{k=0}^t (a_k + c_k + i_{k-1} - b \cdot i_k)\\
&= \sum_{k=0}^t a_k + \sum_{k=0}^t c_k + \sum_{k=0}^t i_{k-1} + \sum_{k=0}^t i_k(-b) \\
&= \sum_{k=0}^t a_k + \sum_{k=0}^t c_k + \sum_{k=0}^{t-1} i_{k} + \sum_{k=0}^t i_k(-b) \\
&= \sum_{k=0}^t a_k + \sum_{k=0}^t c_k + \sum_{k=0}^t i_k(1-b) \\
&= \digsum(a) + \digsum(c) + (1-b)\sum i_k.
\end{align}
Notice we adjusted the indices of the sums for $i_k$ thanks to the fact that $i_{-1} = i_t = 0$.
Since $1-b$ is negative, this is strictly less than $\digsum(a) + \digsum(c)$ iff any $i_k$ is 1.
A: Let $ N= \{0,1,...,n-1\}$, $ I_Z(i) $ be the characteristic function of $ Z \subseteq N $,
$ a=\{ I_A(i)\}_i, $ $ b=\{ I_B(i)\}_i, $ $w(a)=|A|, w(b)=|B|, w(a+b)=|A\cup B|.$  So $  |A\cup B|=|A|+|B| \Leftrightarrow A\cap B=\oslash. $
