Consistency Principle for Binary Mess

Theorem
Let $S$ be a set.

Let $M$ be a binary mess on $S$.

Then there exists a mapping $f : S \to \Bbb B$ such that:
 * $f$ is consistent with $M$

Proof
Let $I$ be the set of all finite subsets of $S$.

Recall that by definition of binary mess:
 * $\ds M \subseteq \bigcup_{P \mathop \in I} \Bbb B^P$

is a set of mappings from finite subsets of $S$ to a Boolean domain.

For every $P \in I$, define:
 * $M_P = \set {t \in M : \Dom t = P}$

By Binary Mess Axiom $(\text M 2)$ we have that $M_P$ cannot be empty.

For every $J \subseteq I$, define:
 * $Z_J = \set {z \in M^J : \paren {\forall P \in \Dom z : \map z P \in M_P} \land \paren {\forall P, Q \in \Dom z : t_1 {\restriction_R} = t_2 {\restriction_R} } }$

where:
 * $t_1 = \map z P$
 * $t_2 = \map z Q$
 * $R = \Dom {t_1} \cap \Dom {t_2}$
 * $M^J$ denotes the set of all mappings from $J$ to $M$.

Define:
 * $\ds Z = \bigcup_{J \mathop \subseteq I} Z_J$
 * $X_P = \set {z \in Z : P \in \Dom z}$
 * $F = \set {X_P : P \in I}$

Non-Emptiness
By Empty Set is Subset of All Sets and Cardinality of Empty Set:
 * $\O \in I$

Therefore:
 * $X_\O \in F$

Thus:
 * $F \ne \O$

Additionally:
 * $\set \O \in J$

Thus:
 * $z_\O \in Z_{\set \O}$

where:
 * $\map z \O = t_\O$

and $t_\O$ is the empty mapping.

Note that $t_\O \in M$ by definition of binary mess and Empty Mapping is Unique.

Therefore, $z_\O \in Z$ and:
 * $Z \ne \O$

Finite Intersection Property
Let $\set {X_1, \dotsc, X_k} \subset F$ be a non-empty finite subset of $F$.

By definition of $F$, for each $1 \le i \le k$, there is some $P \in I$ such that $X_i = X_P$.

By the Principle of Finite Choice, let:
 * $\set {P_1, \dotsc, P_k}$

be such that, for all $1 \le i \le k$:
 * $X_i = X_{P_i}$

Define:
 * $\ds Q = \bigcup_{i \mathop = 1}^k P_i$

By definition of binary mess, there is some $t \in M$ such that:
 * $\Dom t = Q$

Define $z : \set {P_1, \dotsc, P_k} \to M$ as:
 * $\map z P = t {\restriction_P}$

As every $P_i$ is a finite subset of $S$, by definition of binary mess:
 * $t {\restriction_{P_i} } \in M$

Additionally, as $\Dom t = Q \supseteq P_i$ for every $i$, it follows that:
 * $\Dom {t {\restriction_{P_i} } } = P_i$

Thus, for every $1 \le i \le k$:
 * $\map z {P_i} \in M_{P_i}$

Furthermore, for every $i, j$:
 * $\paren {t {\restriction_{P_i} } } {\restriction_R} = t {\restriction_R} = \paren {t {\restriction_{P_j} } } {\restriction_R}$

where $R = P_i \cap P_j$.

Thus, $z$ satisfies all of the requirements so that:
 * $z \in Z_{\set P_1, \dotsc, P_k}$

Hence:
 * $z \in Z$

As every $P_i \in \Dom z$:
 * $\ds z \in \bigcap_{i \mathop = 1}^k X_{P_i} = \bigcap_{i \mathop = 1}^k X_i$

Therefore, $F$ satisfies the finite intersection property.

Ultrafilter
By the, there exists an ultrafilter $U$ on $Z$ such that:
 * $U \supseteq F$

Let $P \in I$ be arbitrary.

For every $t \in M_P$, define:
 * $Y_t = \set {z \in X_P : \map z P = t}$

Let $t, t' \in M_P$ such that:
 * $Y_t \in U \land Y_{t'} \in U$

By definition of filter:
 * $Y_t \cap Y_{t'} \in U$

But then, as $\O \notin U$ by definition:
 * $\exists z \in Y_t \cap Y_{t'}$

Thus:
 * $t = \map z P = t'$

Now, suppose there is no $t \in M_P$ such that:
 * $Y_t \in U$

Then, by definition of ultrafilter, for every $t \in M_P$:
 * $\relcomp Z {Y_t} = \set {z \in Z : P \notin \Dom z \lor \map z P \ne t}$

As:
 * $M_P \subseteq \Bbb B^P$

it follows, by:
 * Set of Mappings between Finite Sets is Finite
 * Subset of Finite Set is Finite

that $M_P$ is a finite set.

Thus:
 * $N = \set {M_P} \cup \set {\relcomp Z {Y_t} : t \in M_P} \subseteq U$

is also finite.

Consider an arbitrary $z \in Z$.
 * If $P \notin \Dom z$, then $z \notin M_P$
 * Otherwise, there exists some $t \in M_P$ such that $\map z P = t$
 * But then, $z \notin Y_t$

Thus, for every $z \in Z$:
 * $z \notin N$

Hence:
 * $\ds \bigcap N = \O$

contradicting the finite intersection property in the definition of ultrafilter.

Thus, for every $P \in I$, there is a unique $t_P \in M_P$ such that $Y_{t_P} \in U$.

Consistent Mapping
Define $f \subseteq S \times \Bbb B$ as:
 * $\ds f = \bigcup_{P \mathop \in I} t_P$

For every $x \in S$:
 * $\Dom {t_{\set x} } = \set x$

Thus:
 * $\tuple {x, \map {t_{\set x} } x} \in f$

and $f$ is left-total by definition.

Now, suppose that, for $x \in S$:
 * $\tuple {x, b_1} \in f \land \tuple {x, b_2} \in f$

By definition of $f$, there are $P, Q \in I$ such that:
 * $\tuple {x, b_1} \in t_P \land \tuple {x, b_2} \in t_Q$

By definition of $t_P$:
 * $Y_{t_P} \in U \land Y_{t_Q} \in U$

Thus, by definition of filter:
 * $Y_{t_P} \cap Y_{t_Q} \in U$

But as $\O \notin U$ by definition of filter:
 * $\exists z \in Y_{t_P} \cap Y_{t_Q}$

That is:
 * $\exists z \in Z : \set {P, Q} \subseteq \Dom z \land \map z P = t_P \land \map z Q = t_Q$

Equivalently, there exists some $J \subseteq I$ such that:
 * $\set {P, Q} \subseteq J$
 * $\exists z \in Z_J : \map z P = t_P \land \map z Q = t_Q$

But by definition of $Z_J$:
 * $t_P {\restriction_{P \mathop \cap Q} } = t_Q {\restriction_{P \mathop \cap Q} }$

As $x \in \Dom {t_P} \cap \Dom {t_Q} = P \cap Q$:
 * $\map {t_P {\restriction_{P \mathop \cap Q} } } x = \map {t_Q {\restriction_{P \mathop \cap Q} } } x$

Or, in other words:
 * $b_1 = b_2$

Thus, $f$ is many-to-one by definition.

Hence, by definition, $f$ is a mapping.

Let $P \in I$ be arbitrary.

By definition of $f$:
 * $t_P \subseteq f$

By Subset of Many-to-One Relation is Restriction, and $\Dom {t_P} = P$:
 * $t_P = f {\restriction_P}$

As $t_P \in M_P \subseteq M$:
 * $f {\restriction_P} \in M$

Therefore, $f$ is consistent with $M$.