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         S(T, P) = (\emptyset, \emptyset) \textrm{   (fixpoint reached)}
     \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 ${\sf NP}$