Let $S = \{1, 2, ..., 6\}$. We want to find the number of functions $F$ that map subsets of $S$ to subsets of $S$, such that for any subsets $A, B \subseteq S$, we have $F(F(A) \cup B) = A \cap F(B)$.

1. Problem Description

Let $S = \{1, 2, ..., 6\}$. We want to find the number of functions $F$ that map subsets of $S$ to subsets of $S$, such that for any subsets $A, B \subseteq S$, we have $F(F(A) \cup B) = A \cap F(B)$.

2. Solution Steps

Let $A = \emptyset$. Then the given condition becomes

$F(F(\emptyset) \cup B) = \emptyset \cap F(B) = \emptyset$.

This means that $F(\emptyset) \cup B$ is always mapped to $\emptyset$ by $F$. Let $C = F(\emptyset) \cup B$.

$F(C) = \emptyset$.

If $B = \emptyset$, then $C = F(\emptyset)$, and we get $F(F(\emptyset)) = \emptyset$.

Let $B = S$. Then $F(\emptyset) \cup S = S$, which implies $F(S) = \emptyset$.

Let $A = S$. Then $F(F(S) \cup B) = S \cap F(B)$.

Since $F(S) = \emptyset$, we have $F(\emptyset \cup B) = F(B) = S \cap F(B) = F(B)$.

This doesn't give us new information.

Let $B = \emptyset$. Then $F(F(A) \cup \emptyset) = F(F(A)) = A \cap F(\emptyset)$.

We also know that $F(F(\emptyset)) = \emptyset$.

Let $A = F(\emptyset)$. Then $F(F(F(\emptyset))) = F(\emptyset) \cap F(\emptyset)$.

So $F(\emptyset) = F(\emptyset) \cap F(\emptyset) = F(\emptyset)$, which is trivial.

Let $A=S$. Then $F(F(S) \cup B) = F(\emptyset \cup B) = F(B) = S \cap F(B) = F(B)$. This is also trivial.

Consider $F(F(A) \cup B) = A \cap F(B)$.

Let $A = B$. Then $F(F(A) \cup A) = A \cap F(A)$.

If $F(A) \subseteq A$, then $F(A) \cup A = A$, and $F(F(A) \cup A) = F(A)$. So $F(A) = A \cap F(A) = F(A)$.

However, if $A \subseteq F(A)$, then $F(A) \cup A = F(A)$, and $F(F(A) \cup A) = F(F(A))$.

So $F(F(A)) = A \cap F(A) = A$.

We also know that $F(S) = \emptyset$.

Let $A=B=S$. Then $F(F(S)\cup S) = S \cap F(S)$, so $F(\emptyset \cup S) = F(S) = S \cap \emptyset = \emptyset$.

$F(S) = \emptyset$.

We want to find the number of functions $F$.

For each $x \in S$, we either have $x \in F(A)$ or $x \notin F(A)$.

We know that $F(S) = \emptyset$, which implies that $x \notin F(S)$ for all $x \in S$.

We also know that $F(\emptyset)$ is crucial.

Suppose $F(\emptyset) = \{1, 2, ..., k\}$.

Let $x \in S$. For each element $x$, either $x \in F(\emptyset)$ or $x \notin F(\emptyset)$.

If $x \in A$, then $x \in A \cap F(B)$. If $x \notin A$, then $x \notin A \cap F(B)$.

For each $x$, consider $A=\{x\}$, $B = \emptyset$.

Then $F(F(\{x\}) \cup \emptyset) = \{x\} \cap F(\emptyset)$, which implies $F(F(\{x\})) = \{x\} \cap F(\emptyset)$.

If $x \in F(\emptyset)$, then $F(F(\{x\})) = \{x\}$.

If $x \notin F(\emptyset)$, then $F(F(\{x\})) = \emptyset$.

If we have $x \in S$, then $x \in F(\emptyset)$ or $x \notin F(\emptyset)$.

We also have $F(\emptyset) = X \subseteq S$. The number of possible subsets is $2^6 = 64$.

Suppose $F(\emptyset) = A$. Then $F(A) = \emptyset$. So, $F(F(\emptyset)) = F(A) = \emptyset$.

Let $x \in A$. Then $F(F(\{x\})) = \{x\}$. Let $x \notin A$. Then $F(F(\{x\})) = \emptyset$.

This seems to be linked to whether $x$ is in $F(\emptyset)$.

Let $S=\{1\}$. If $A = \emptyset$, $B = \emptyset$, then $F(F(\emptyset) \cup \emptyset) = \emptyset \cap F(\emptyset)$, so $F(F(\emptyset)) = \emptyset$.

If $A = \emptyset$, $B = \{1\}$, then $F(F(\emptyset) \cup \{1\}) = \emptyset \cap F(\{1\})$, so $F(F(\emptyset) \cup \{1\}) = \emptyset$.

If $A = \{1\}$, $B = \emptyset$, then $F(F(\{1\}) \cup \emptyset) = \{1\} \cap F(\emptyset)$, so $F(F(\{1\})) = \{1\} \cap F(\emptyset)$.

If $A = \{1\}$, $B = \{1\}$, then $F(F(\{1\}) \cup \{1\}) = \{1\} \cap F(\{1\})$.

There are $2^6=64$ choices for $F(\emptyset)$. For each $A \subseteq S$, we must have $F(A)$.

Consider the case $S = \{1\}$. $F$ maps subsets of $S$ to subsets of $S$.

Subsets of $S$ are $\emptyset$ and $\{1\}$.

So, we need to find how many functions satisfy $F(F(A) \cup B) = A \cap F(B)$ for all $A, B \subseteq S$.

We have the following cases for $F$:

$F(\emptyset) = \emptyset$, $F(\{1\}) = \emptyset$.

$F(\emptyset) = \emptyset$, $F(\{1\}) = \{1\}$.

$F(\emptyset) = \{1\}$, $F(\{1\}) = \emptyset$.

$F(\emptyset) = \{1\}$, $F(\{1\}) = \{1\}$.

3. Final Answer

64