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In a comment Peter LeFanu Lumsdaine has conjectured a classification, which shows that there are $16$ (!) binary subset operations. They can be combined from $1 \cdot 2 \cdot 2 \cdot 4$ independent cases. For reference, let me write them down:

• $\alpha(\emptyset,\emptyset)$ must be $\emptyset$

• For non-empty sets $U$, $\alpha(U,\emptyset)$ can be one of $\emptyset$, $U$

• For non-empty sets $V$, $\alpha(\emptyset,V)$ can be one of $\emptyset$, $V$

• For non-empty sets $U,V$, $\alpha(U,V)$ can be one of $\emptyset, U, V, U \cup V$.

In a comment Peter LeFanu Lumsdaine has conjectured a classification, which shows that there are $16$ (!) binary subset operations. They can be combined from $1 \cdot 2 \cdot 2 \cdot 4$ independent cases. For reference, let me write them down:

• $\alpha(\emptyset,\emptyset)$ must be $\emptyset$

• For non-empty sets $U$, $\alpha(U,\emptyset)$ can be one of $\emptyset$, $U$

• For non-empty sets $V$, $\alpha(\emptyset,V)$ can be one of $\emptyset$, $V$

• For non-empty sets $U,V$, $\alpha(U,V)$ can be one of $\emptyset, U, V, U \cup V$.

Update. According to the now proven classification in terms of deflationary maps, let me also mention the corresponding $16$ deflationary maps $M : P([2]) \to P([2])$:

• $M(\emptyset)$ must be $\emptyset$
• $M(\{1\})$ can be one of $\emptyset, \{1\}$
• $M(\{2\})$ can be one of $\emptyset, \{2\}$
• $M(\{1,2\})$ can be one of $\emptyset, \{1\}, \{2\}, \{1,2\}$

In a comment Peter has conjectured a classification, which shows that there are $16$ (!) binary subset operations. They can be combined from $1 \cdot 2 \cdot 2 \cdot 4$ independent cases. For reference, let me write them down:

• $\alpha(\emptyset,\emptyset)$ must be $\emptyset$

• For non-empty sets $U$, $\alpha(U,\emptyset)$ can be one of $\emptyset$, $U$

• For non-empty sets $V$, $\alpha(\emptyset,V)$ can be one of $\emptyset$, $V$

• For non-empty sets $U,V$, $\alpha(U,V)$ can be one of $\emptyset, U, V, U \cup V$.

In a comment Peter LeFanu Lumsdaine has conjectured a classification, which shows that there are $16$ (!) binary subset operations. They can be combined from $1 \cdot 2 \cdot 2 \cdot 4$ independent cases. For reference, let me write them down:

• $\alpha(\emptyset,\emptyset)$ must be $\emptyset$

• For non-empty sets $U$, $\alpha(U,\emptyset)$ can be one of $\emptyset$, $U$

• For non-empty sets $V$, $\alpha(\emptyset,V)$ can be one of $\emptyset$, $V$

• For non-empty sets $U,V$, $\alpha(U,V)$ can be one of $\emptyset, U, V, U \cup V$.

In a comment Peter has conjectured a classification, which shows that there are $16$ (!) binary subset operations. Here they are. These expression have been generated using ChatGPT, so need some verification. It's not a proper answer, hence CW, but I also didn't want to add these to the post itself to reduce scrolling.$ \newcommand\template[4]{\begin{cases} #1 & \text{if both $U_1 = \emptyset$ and $U_2 = \emptyset$} \\ #2 & \text{if $U_1 \ne \emptyset$ and $U_2 = \emptyset$} \\ #3 & \text{if $U_1 = \emptyset$ and $U_2 \ne \emptyset$} \\ #4 & \text{if $U_1 \ne \emptyset$ and $U_2 \ne \emptyset$.} \end{cases}} \newcommand\e{\emptyset}$

1. $\emptyset$

2. $\template\e\e\e{U_1}$

3. $\template\e\e\e{U_2}$

4. $\template\e\e\e{U_1 \cup U_2}$

5. $\template\e\e{U_2}\e$

6. $\template\e\e{U_2}{U_1}$

7. $\template\e\e{U_2}{U_2}$

8. $\template\e\e{U_2}{U_1 \cup U_2}$

9. $\template\e{U_1}\e\e$

10. $\template\e{U_1}\e{U_1}$

11. $\template\e{U_1}\e{U_2}$

12. $\template\e{U_1}\e{U_1 \cup U_2}$

13. $\template\e{U_1}{U_2}\e$

14. $\template\e{U_1}{U_2}{U_1}$

15. $\template\e{U_1}{U_2}{U_2}$

16. $\template\e{U_1}{U_2}{U_1 \cup U_2}$

In a comment Peter has conjectured a classification, which shows that there are $16$ (!) binary subset operations. They can be combined from $1 \cdot 2 \cdot 2 \cdot 4$ independent cases. For reference, let me write them down:

• $\alpha(\emptyset,\emptyset)$ must be $\emptyset$

• For non-empty sets $U$, $\alpha(U,\emptyset)$ can be one of $\emptyset$, $U$

• For non-empty sets $V$, $\alpha(\emptyset,V)$ can be one of $\emptyset$, $V$

• For non-empty sets $U,V$, $\alpha(U,V)$ can be one of $\emptyset, U, V, U \cup V$.