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Spencer Killen 2023-05-17 21:20:01 -06:00
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\end{align*}
\begin{proposition}\label{}
Given a poset $\langle \mathcal{L}, \leq \rangle$,
an operator $o: \mathcal{L} \rightarrow \mathcal{L}$ is $\leq$-monotone iff it is $\geq$-monotone
\end{proposition}
\begin{proof}
($\Rightarrow$)
Initially, we have $\forall x, y \in \mathcal{L}: x \leq y \implies o(x) \leq o(y)$.
Let $a, b \in \mathcal{L}$ such that $a \geq b$. We have $b \leq a$ and can apply our initial assumption to get $o(b) \leq o(a)$.
This gives us $o(a) \geq o(b)$.
We can generalize to obtain $\forall x, y \in \mathcal{L}: x \geq y \implies o(x) \geq o(y)$.
($\Leftarrow$) Proof is more or less the same
\end{proof}
% Let $\leq$ and $\leq_i$ be the orderings over the set $\mathcal{L}^2$ such that for each ${(T, P), (X, Y) \in \mathcal{L}^2}$
% \begin{align*}
% (a, b) \leq (x, y) &\textit{ iff } a \leq x \textrm{ and } b \leq y\\
% (a, b) \leq_i (x, y) &\textit{ iff } a \leq x \textrm{ and } \boxed{y} \leq b
% \end{align*}
% \begin{proposition}
% Given a poset $\langle \mathcal{L}^2, \leq \rangle$, an operator $o: \wp(\mathcal{L})^2 \rightarrow \wp(\mathcal{L})^2$ is $\leq$-monotone iff it is $\leq_i$-monotone
% \end{proposition}
% \begin{proof}
% ($\Rightarrow$) Initially, we have ${\forall (x, y), (a, b) \in \mathcal{L}^2: (x, y) \leq (a, b) \implies o(x, y) \leq o(a, b)}$.
% If we rearrange the variables we get
% \begin{align*}
% &\forall x, a \in \mathcal{L}: \forall y, b \in \mathcal{L}:\\
% &~~~~~(x \leq a \land y \leq b) \implies ((o_1(x, y) \leq o_1(a, b)) \land (o_2(x, y) \leq o_2(a, b)))
% \end{align*}
% Let $(u, v), (i, k) \in \mathcal{L}^2$ such that $(u, v) \leq_i (i, k)$. We have $(u, k) \leq (i, v)$. We can apply the initial assumption to obtain
% $o(u, k) \leq o(i, v)$. This is equivalent to
% \begin{align*}
% (o_1(u, k) \leq o_1(i, v)) \land (o_2(u, k) \leq o_2(i, v))
% \end{align*}
% which can be rewritten as
% \begin{align*}
% (o_1(u, k) \leq o_1(i, v)) \land \boxed{(o_2(i, v)) \leq o_2(u, k))}
% \end{align*}
% \end{proof}
\begin{proposition}
An operator $A : L^2 \rightarrow L^2$ is symmetric and monotone
with respect to both $\leq_i$ and $\leq$ if and only if there is a monotone operator
$O : L \rightarrow L$ such that for every $x, y \in L, A(x, y) = (O (x), O (y ))$
\end{proposition}
\begin{proof}
($\Rightarrow$)
From Proposition 5 and 6 we have for any $x \in L$, $A_1(\cdot, x)$ and $A_1(x, \cdot)$ are monotone and $A_1(x, \cdot)$ is antimonotone.
By Proposition 2, $A_1(x, \cdot)$ is constant, denote this constant as the function $O(x)$.
By the symmetric condition, we have $A_1(x, \cdot) = A_2(\cdot, x)$, thus $A(x, y) = (O(x), O(y))$. It follows from the monotonicity of $A$ that $O$ is monotone.
($\Leftarrow$)
Clearly $A$ is symmetric, and $\leq$-monotone (Given that $O$ is $\leq$-monotone). Using proposition 3.1 (above) $O(x)$ is $\geq$-monotone, thus $A$ is $\leq_i$-monotone as well.
\end{proof}
\section{The Polynomial Heirarchy}
Intuitive definitions of ${\sf NP}$
\begin{itemize}