Paper/Paper.thy
changeset 2502 074304d56eb2
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   781 
   781 
   782   \begin{thm}[Support of Abstractions]\label{suppabs} 
   782   \begin{thm}[Support of Abstractions]\label{suppabs} 
   783   Assuming @{text x} has finite support, then\\[-6mm] 
   783   Assuming @{text x} has finite support, then\\[-6mm] 
   784   \begin{center}
   784   \begin{center}
   785   \begin{tabular}{l@ {\hspace{2mm}}c@ {\hspace{2mm}}l}
   785   \begin{tabular}{l@ {\hspace{2mm}}c@ {\hspace{2mm}}l}
   786   @{thm (lhs) supp_abs(1)[no_vars]} & $=$ & @{thm (rhs) supp_abs(1)[no_vars]}\\
   786   @{thm (lhs) supp_Abs(1)[no_vars]} & $=$ & @{thm (rhs) supp_Abs(1)[no_vars]}\\
   787   @{thm (lhs) supp_abs(2)[no_vars]} & $=$ & @{thm (rhs) supp_abs(2)[no_vars]}\\
   787   @{thm (lhs) supp_Abs(2)[no_vars]} & $=$ & @{thm (rhs) supp_Abs(2)[no_vars]}\\
   788   @{thm (lhs) supp_abs(3)[where bs="as", no_vars]} & $=$ & @{thm (rhs) supp_abs(3)[where bs="as", no_vars]}
   788   @{thm (lhs) supp_Abs(3)[where bs="as", no_vars]} & $=$ & @{thm (rhs) supp_Abs(3)[where bs="as", no_vars]}
   789   \end{tabular}
   789   \end{tabular}
   790   \end{center}
   790   \end{center}
   791   \end{thm}
   791   \end{thm}
   792 
   792 
   793   \noindent
   793   \noindent
   794   Below we will show the first equation. The others 
   794   Below we will show the first equation. The others 
   795   follow by similar arguments. By definition of the abstraction type @{text "abs_set"} 
   795   follow by similar arguments. By definition of the abstraction type @{text "abs_set"} 
   796   we have 
   796   we have 
   797   %
   797   %
   798   \begin{equation}\label{abseqiff}
   798   \begin{equation}\label{abseqiff}
   799   @{thm (lhs) abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} \;\;\text{if and only if}\;\; 
   799   @{thm (lhs) Abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} \;\;\text{if and only if}\;\; 
   800   @{thm (rhs) abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]}
   800   @{thm (rhs) Abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]}
   801   \end{equation}
   801   \end{equation}
   802 
   802 
   803   \noindent
   803   \noindent
   804   and also
   804   and also
   805   %
   805   %
   812   \eqref{permute} and $\alpha$-equivalence being equivariant 
   812   \eqref{permute} and $\alpha$-equivalence being equivariant 
   813   (see Lemma~\ref{alphaeq}). With these two facts at our disposal, we can show 
   813   (see Lemma~\ref{alphaeq}). With these two facts at our disposal, we can show 
   814   the following lemma about swapping two atoms in an abstraction.
   814   the following lemma about swapping two atoms in an abstraction.
   815   
   815   
   816   \begin{lemma}
   816   \begin{lemma}
   817   @{thm[mode=IfThen] abs_swap1(1)[where bs="as", no_vars]}
   817   @{thm[mode=IfThen] Abs_swap1(1)[where bs="as", no_vars]}
   818   \end{lemma}
   818   \end{lemma}
   819 
   819 
   820   \begin{proof}
   820   \begin{proof}
   821   This lemma is straightforward using \eqref{abseqiff} and observing that
   821   This lemma is straightforward using \eqref{abseqiff} and observing that
   822   the assumptions give us @{term "(a \<rightleftharpoons> b) \<bullet> (supp x - as) = (supp x - as)"}.
   822   the assumptions give us @{term "(a \<rightleftharpoons> b) \<bullet> (supp x - as) = (supp x - as)"}.
   826   \noindent
   826   \noindent
   827   Assuming that @{text "x"} has finite support, this lemma together 
   827   Assuming that @{text "x"} has finite support, this lemma together 
   828   with \eqref{absperm} allows us to show
   828   with \eqref{absperm} allows us to show
   829   %
   829   %
   830   \begin{equation}\label{halfone}
   830   \begin{equation}\label{halfone}
   831   @{thm abs_supports(1)[no_vars]}
   831   @{thm Abs_supports(1)[no_vars]}
   832   \end{equation}
   832   \end{equation}
   833   
   833   
   834   \noindent
   834   \noindent
   835   which by Property~\ref{supportsprop} gives us ``one half'' of
   835   which by Property~\ref{supportsprop} gives us ``one half'' of
   836   Theorem~\ref{suppabs}. The ``other half'' is a bit more involved. To establish 
   836   Theorem~\ref{suppabs}. The ``other half'' is a bit more involved. To establish 
   854   \noindent
   854   \noindent
   855   using \eqref{suppfun}. Assuming @{term "supp x - as"} is a finite set,
   855   using \eqref{suppfun}. Assuming @{term "supp x - as"} is a finite set,
   856   we further obtain
   856   we further obtain
   857   %
   857   %
   858   \begin{equation}\label{halftwo}
   858   \begin{equation}\label{halftwo}
   859   @{thm (concl) supp_abs_subset1(1)[no_vars]}
   859   @{thm (concl) Abs_supp_subset1(1)[no_vars]}
   860   \end{equation}
   860   \end{equation}
   861 
   861 
   862   \noindent
   862   \noindent
   863   since for finite sets of atoms, @{text "bs"}, we have 
   863   since for finite sets of atoms, @{text "bs"}, we have 
   864   @{thm (concl) supp_finite_atom_set[where S="bs", no_vars]}.
   864   @{thm (concl) supp_finite_atom_set[where S="bs", no_vars]}.