Monotonically normal implies normal

Statement
Any monotonically normal space is a normal space.

Monotonically normal space
A topological space $$X$$ is termed monotonically normal if it is a $$T_1$$ space and there exists an operator (called a monotone normality operator) $$G$$ from pairs of disjoint closed subsets $$(A,B)$$ to open subsets such that:


 * For all disjoint closed subsets $$A,B$$, $$G(A,B)$$ is an open subset containing $$A$$ and whose closure is disjoint from $$B$$.
 * For closed subsets $$A,A',B,B'$$, if $$A \subseteq A'$$, $$B' \subseteq B$$, with $$A,B$$ disjoint and $$A',B'$$ disjoint, we have $$G(A,B) \subseteq G(A',B')$$.

Normal space
A topological space $$X$$ is termed normal if it is a $$T_1$$ space and, for any two disjoint closed subsets $$A$$ and $$B$$, there exist disjoint open subsets $$U$$ and $$V$$ such that $$U$$ contains $$A$$ and $$V$$ contains $$B$$.

Proof
Given: A monotonically normal space $$X$$ with a monotone normality operator $$G$$.

To prove: $$X$$ is a normal space.

Proof: The $$T_1$$-space part follows by definition. For the other part, note that $$U = G(A,B)$$ and $$V = X \setminus \overline{G(A,B)}$$ are disjoint open subsets containing $$A$$ and $$B$$ respectively.