Let $a$ and $b$ be (not necessarily proper) subsets of $\mathbb{Z}$. Let the operator $\oplus$ be defined recursively as follows:
$$a \oplus ∅ = a$$ $$a \oplus b = (a△b) \oplus (a \cap b)^*$$
Being $a△b$ the symmetric difference between $a$ and $b$, and $a^* ≔ \{z∈Z:z-1 \in a\}$. Think of $a$ and $b$ as bit strings, where each element identifies a position, e.g. $\{2,0,-1\}$ represents ${101.1}_2={5.5}_{10}$. Then the operator $\oplus$ is the addition with $(a \cap b)^*$ being the carried bits shifted 1 bit to the left.
If both $a$ and $b$ have a maximum element, then I think I can show that for any such $a$ and $b$ the recursion will eventually halt, and the result will also have a maximum element.
But what happens if $a$ has no maximum element, e.g. when $a= \mathbb{N}$?
Let $b=\{0\}$. Carrying out the first steps of the recursion yields: $$\mathbb{N} \oplus \{0\}=(\mathbb{N} \setminus \{0\})\oplus \{0\}^*=(\mathbb{N} \setminus \{0\})\oplus\{1\}$$ $$=(\mathbb{N}\setminus\{0,1\}) \oplus \{1\}^*=(\mathbb{N}\setminus\{0,1\}) \oplus \{2\}$$ $$=(\mathbb{N}\setminus\{0,1,2\}) \oplus \{2\}^*=(\mathbb{N}\setminus\{0,1,2\}) \oplus \{3\}$$
Intuitionally, I would expect that after an infinite number of steps (¿after ω steps?), we would only be left with the empty set as we keep on removing elements from N.
Is there any way to argue that indeed $\mathbb{N} \oplus \{0\}=\emptyset$?
Is there any “framework” in which $\mathbb{N} \oplus \{0\}=\emptyset$?
Is the operator $\oplus$ even well defined?