# HG changeset patch # User Cezary Kaliszyk # Date 1285915490 -32400 # Node ID 074304d56eb2e054075481d81bddc640c0e74a5c # Parent 6abeee9e52e326f9c01bf306870d823064a17e9e Made the paper to compile with the renamings. diff -r 6abeee9e52e3 -r 074304d56eb2 Paper/Paper.thy --- a/Paper/Paper.thy Wed Sep 29 09:51:57 2010 -0400 +++ b/Paper/Paper.thy Fri Oct 01 15:44:50 2010 +0900 @@ -783,9 +783,9 @@ Assuming @{text x} has finite support, then\\[-6mm] \begin{center} \begin{tabular}{l@ {\hspace{2mm}}c@ {\hspace{2mm}}l} - @{thm (lhs) supp_abs(1)[no_vars]} & $=$ & @{thm (rhs) supp_abs(1)[no_vars]}\\ - @{thm (lhs) supp_abs(2)[no_vars]} & $=$ & @{thm (rhs) supp_abs(2)[no_vars]}\\ - @{thm (lhs) supp_abs(3)[where bs="as", no_vars]} & $=$ & @{thm (rhs) supp_abs(3)[where bs="as", no_vars]} + @{thm (lhs) supp_Abs(1)[no_vars]} & $=$ & @{thm (rhs) supp_Abs(1)[no_vars]}\\ + @{thm (lhs) supp_Abs(2)[no_vars]} & $=$ & @{thm (rhs) supp_Abs(2)[no_vars]}\\ + @{thm (lhs) supp_Abs(3)[where bs="as", no_vars]} & $=$ & @{thm (rhs) supp_Abs(3)[where bs="as", no_vars]} \end{tabular} \end{center} \end{thm} @@ -796,8 +796,8 @@ we have % \begin{equation}\label{abseqiff} - @{thm (lhs) abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} \;\;\text{if and only if}\;\; - @{thm (rhs) abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} + @{thm (lhs) Abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} \;\;\text{if and only if}\;\; + @{thm (rhs) Abs_eq_iff(1)[where bs="as" and cs="bs", no_vars]} \end{equation} \noindent @@ -814,7 +814,7 @@ the following lemma about swapping two atoms in an abstraction. \begin{lemma} - @{thm[mode=IfThen] abs_swap1(1)[where bs="as", no_vars]} + @{thm[mode=IfThen] Abs_swap1(1)[where bs="as", no_vars]} \end{lemma} \begin{proof} @@ -828,7 +828,7 @@ with \eqref{absperm} allows us to show % \begin{equation}\label{halfone} - @{thm abs_supports(1)[no_vars]} + @{thm Abs_supports(1)[no_vars]} \end{equation} \noindent @@ -856,7 +856,7 @@ we further obtain % \begin{equation}\label{halftwo} - @{thm (concl) supp_abs_subset1(1)[no_vars]} + @{thm (concl) Abs_supp_subset1(1)[no_vars]} \end{equation} \noindent