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@ -2,3 +2,5 @@
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\renewcommand{\L}{\mathcal{L}}
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\renewcommand{\L}{\mathcal{L}}
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\newcommand{\lte}{\preceq}
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\newcommand{\lte}{\preceq}
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\newcommand{\image}[1]{[#1]}
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\newcommand{\image}[1]{[#1]}
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\newcommand{\define}{\coloneqq}
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\newcommand{\union}{\cup}
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18
note.tex
18
note.tex
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@ -5,6 +5,8 @@
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\usepackage{bm}
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\usepackage{bm}
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\usepackage[english]{babel}
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\usepackage[english]{babel}
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\usepackage{amsthm}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{mathtools}
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\newcommand{\jh}[1]{{\leavevmode\color{blue!50!red}#1}}
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\newcommand{\jh}[1]{{\leavevmode\color{blue!50!red}#1}}
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\input{notation.tex}
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\input{notation.tex}
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@ -36,7 +38,21 @@ $$\fixpointsOf(S)$$
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For a \Monotone function $o$ over a \CompleteLattice $\langle \L, \lte \rangle$, we have that $\langle o\imageNoLink{\L}, \lte \rangle$ is a \CompleteLattice.
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For a \Monotone function $o$ over a \CompleteLattice $\langle \L, \lte \rangle$, we have that $\langle o\imageNoLink{\L}, \lte \rangle$ is a \CompleteLattice.
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{proof}
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foo
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Consider $\langle \L', \lte' \rangle$ where
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\begin{align*}
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\L' \define \{ x' ~|~ x \in \L \} \\
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x' \lte' y' \iff x \lte y \textrm{ where } x, y \in \L
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\end{align*}
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Clearly, $\langle \L', \lte' \rangle$ is a \CompleteLattice.
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Combining $\L$ and $\L'$
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\begin{align*}
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\L^* \define \{ \top^*, \bot^* \} \union \L \union \L' \\
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\forall x^*, y^* \in \L^*, x^* \lte y^* \iff \begin{cases}
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(x^* = \bot^*) \lor (y^* = \top^*) \\
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two \\
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three
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\end{cases}
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\end{align*}
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\end{proof}
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\end{proof}
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\bibliographystyle{plain}
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\bibliographystyle{plain}
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