(*<*)theory Slides3imports "~~/src/HOL/Library/LaTeXsugar" "Nominal"begindeclare [[show_question_marks = false]]notation (latex output) set ("_") and Cons ("_::/_" [66,65] 65) (*>*)text_raw {* \renewcommand{\slidecaption}{UNIF, Edinburgh, 14.~July 2010} \newcommand{\abst}[2]{#1.#2}% atom-abstraction \newcommand{\pair}[2]{\langle #1,#2\rangle} % pairing \newcommand{\susp}{{\boldsymbol{\cdot}}}% for suspensions \newcommand{\unit}{\langle\rangle}% unit \newcommand{\app}[2]{#1\,#2}% application \newcommand{\eqprob}{\mathrel{{\approx}?}} \newcommand{\freshprob}{\mathrel{\#?}} \newcommand{\redu}[1]{\stackrel{#1}{\Longrightarrow}}% reduction \newcommand{\id}{\varepsilon}% identity substitution \pgfdeclareradialshading{smallbluesphere}{\pgfpoint{0.5mm}{0.5mm}}% {rgb(0mm)=(0,0,0.9); rgb(0.9mm)=(0,0,0.7); rgb(1.3mm)=(0,0,0.5); rgb(1.4mm)=(1,1,1)} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Quiz} Assuming that \smath{a} and \smath{b} are distinct variables,\\ is it possible to find $\lambda$-terms \smath{M_1} to \smath{M_7} that make the following pairs \alert{$\alpha$-equivalent}? \begin{tabular}{@ {\hspace{14mm}}p{12cm}} \begin{itemize} \item \smath{\lambda a.\lambda b. (M_1\,b)\;} and \smath{\lambda b.\lambda a. (a\,M_1)\;} \item \smath{\lambda a.\lambda b. (M_2\,b)\;} and \smath{\lambda b.\lambda a. (a\,M_3)\;} \item \smath{\lambda a.\lambda b. (b\,M_4)\;} and \smath{\lambda b.\lambda a. (a\,M_5)\;} \item \smath{\lambda a.\lambda b. (b\,M_6)\;} and \smath{\lambda a.\lambda a. (a\,M_7)\;} \end{itemize} \end{tabular} If there is one solution for a pair, can you describe all its solutions? \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[t] \frametitle{% \begin{tabular}{@ {\hspace{-3mm}}c@ {}} \\ \huge Nominal Unification\\[-2mm] \Large Hitting a Sweet Spot\\[5mm] \end{tabular}} \begin{center} Christian Urban \end{center} \begin{center} \small initial spark from Roy Dyckhoff in November 2001\\[0mm] \small joint work with Andy Pitts and Jamie Gabbay\\[0mm] \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-4>[c] \frametitle{One Motivation} \onslide<2->{Typing implemented in Prolog \textcolor{darkgray}{(from a textbook)}}\bigskip\\ \onslide<3->{\color{darkgray} \begin{tabular}{l} type (Gamma, var(X), T) :- member (X,T) Gamma.\smallskip\medskip\\ type (Gamma, app(M, N), T') :-\\ \hspace{3cm}type (Gamma, M, arrow(T, T')),\\ \hspace{3cm}type (Gamma, N, T).\smallskip\medskip\\ type (Gamma, lam(X, M), arrow(T, T')) :-\\ \hspace{3cm}type ((X, T)::Gamma, M, T').\smallskip\medskip\\ member X X::Tail.\\ member X Y::Tail :- member X Tail.\\ \end{tabular}} \only<4>{ \begin{textblock}{6}(2.5,2) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{8cm}\raggedright The problem is that \smath{\lambda x.\lambda x. (x\;x)} will have the types \begin{center} \begin{tabular}{l} \smath{T\rightarrow (T\rightarrow S) \rightarrow S} and\\ \smath{(T\rightarrow S)\rightarrow T \rightarrow S}\\ \end{tabular} \end{center} \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Higher-Order Unification} State of the art at the time: \begin{itemize} \item Lambda Prolog with full Higher-Order Unification\\ \textcolor{darkgray}{(no mgus, undecidable, modulo $\alpha\beta$)}\bigskip \item Higher-Order Pattern Unification\\ \textcolor{darkgray}{(has mgus, decidable, some restrictions, modulo $\alpha\beta_0$)} \end{itemize} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-10>[t] \frametitle{Underlying Ideas} \begin{itemize} \item<1-> Unification (\alert{only}) up to $\alpha$ \item<2-> Swappings / Permutations \only<2-5>{ \begin{center} \begin{tabular}{r@ {\hspace{1mm}}l@ {\hspace{12mm}}r@ {\hspace{1mm}}l} \\ \only<2>{\smath{\textcolor{white}{[b\!:=\!a]}}}% \only<3>{\smath{[b\!:=\!a]}}% \only<4-5>{\smath{\alert{\swap{a}{b}\,\act}}} & \onslide<2-5>{\smath{\lambda a.b}} & \only<2>{\smath{\textcolor{white}{[b\!:=\!a]}}}% \only<3>{\smath{[b\!:=\!a]}}% \only<4-5>{\smath{\alert{\swap{a}{b}\,\act}}} & \onslide<2-5>{\smath{\lambda c.b}}\\ \onslide<3-5>{\smath{=}} & \only<3>{\smath{\lambda a.a}}\only<4-5>{\smath{\lambda b.a}} & \onslide<3-5>{\smath{=}} & \only<3>{\smath{\lambda c.a}}\only<4-5>{\smath{\lambda c.a}}\\ \end{tabular} \end{center}\bigskip \onslide<4-5>{ \begin{center} \begin{tikzpicture} \draw (0,0) node[inner sep=0mm,fill=cream, ultra thick, draw=cream] {\begin{minipage}{8cm} \begin{tabular}{r@ {\hspace{3mm}}l} \smath{\swap{a}{b}\act t} $\;\dn$ & \alert{swap} {\bf all} occurrences of\\ & \smath{b} and \smath{a} in \smath{t} \end{tabular} \end{minipage}}; \end{tikzpicture} \end{center}}\bigskip \onslide<5>{ Unlike for \smath{[b\!:=\!a]\act(-)}, for \smath{\swap{a}{b}\act (-)} we do have if \smath{t =_\alpha t'} then \smath{\pi \act t =_\alpha \pi \act t'.}}} \item<6-> Variables (or holes)\bigskip \begin{center} \onslide<7->{\mbox{}\hspace{-25mm}\smath{\lambda x\hspace{-0.5mm}s .}} \onslide<8-9>{\raisebox{-1.7mm}{\huge\smath{(}}}\raisebox{-4mm}{\begin{tikzpicture} \fill[blue] (0, 0) circle (5mm); \end{tikzpicture}} \onslide<8-9>{\smath{y\hspace{-0.5mm}s}{\raisebox{-1.7mm}{\huge\smath{)}}}}\bigskip \end{center} \only<8-9>{\smath{y\hspace{-0.5mm}s} are the parameters the hole can depend on\onslide<9->{, but then you need $\beta_0$-reduction\medskip \begin{center} \smath{(\lambda x. t) y \longrightarrow_{\beta_0} t[x:=y]} \end{center}}} \only<10>{we will record the information about which parameters a hole \alert{\bf cannot} depend on} \end{itemize} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-4>[c] \frametitle{Terms} \begin{tabular}{lll @ {\hspace{10mm}}lll} \onslide<1->{\pgfuseshading{smallbluesphere}} & \onslide<1->{\colorbox{cream}{\smath{\unit}}} & \onslide<1->{Units} & \onslide<2->{\pgfuseshading{smallbluesphere}} & \onslide<2->{\colorbox{cream}{\smath{a}}} & \onslide<2->{Atoms} \\[5mm] \onslide<1->{\pgfuseshading{smallbluesphere}} & \onslide<1->{\colorbox{cream}{\smath{\pair{t}{t'}}}} & \onslide<1->{Pairs} & \onslide<3->{\pgfuseshading{smallbluesphere}} & \onslide<3->{\colorbox{cream}{\smath{\abst{a}{t}}}} & \onslide<3->{Abstractions}\\[5mm] \onslide<1->{\pgfuseshading{smallbluesphere}} & \onslide<1->{\colorbox{cream}{\smath{\app{F}{t}}}} & \onslide<1->{Funct.} & \onslide<4->{\pgfuseshading{smallbluesphere}} & \onslide<4->{\colorbox{cream}{\smath{\pi\susp X}}} & \onslide<4->{Suspensions} \end{tabular} \only<2>{ \begin{textblock}{13}(1.5,12) \small Atoms are constants \textcolor{darkgray}{(infinitely many of them)} \end{textblock}} \only<3>{ \begin{textblock}{13}(1.5,12) \small \smath{\ulcorner \lambda\abst{a}{a}\urcorner \mapsto \text{fn\ }\abst{a}{a}}\\ \small constructions like \smath{\text{fn\ }\abst{X}{X}} are not allowed \end{textblock}} \only<4>{ \begin{textblock}{13}(1.5,12) \small \smath{X} is a variable standing for a term\\ \small \smath{\pi} is an explicit permutation \smath{\swap{a_1}{b_1}\ldots\swap{a_n}{b_n}}, waiting to be applied to the term that is substituted for \smath{X} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3>[c] \frametitle{Permutations} a permutation applied to a term \begin{center} \begin{tabular}{lrcl} \pgfuseshading{smallbluesphere} & \smath{[]\act c} & \smath{\dn} & \smath{c} \\ \pgfuseshading{smallbluesphere} & \smath{\swap{a}{b}\!::\!\pi\act c} & \smath{\dn} & \smath{\begin{cases} a & \text{if}\;\pi\act c = b\\ b & \text{if}\;\pi\act c = a\\ \pi\act c & \text{otherwise} \end{cases}}\\ \onslide<2->{\pgfuseshading{smallbluesphere}} & \onslide<2->{\smath{\pi\act\abst{a}{t}}} & \onslide<2->{\smath{\dn}} & \onslide<2->{\smath{\abst{\pi\act a}{\pi\act t}}}\\ \onslide<3->{\pgfuseshading{smallbluesphere}} & \onslide<3->{\smath{\pi\act\pi'\act X}} & \onslide<3->{\smath{\dn}} & \onslide<3->{\smath{(\pi @ \pi')\act X}}\\ \end{tabular} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3>[c] \frametitle{Freshness Constraints} Recall \smath{\lambda a. \raisebox{-0.7mm}{\tikz \fill[blue] (0, 0) circle (2.5mm);}} \bigskip\pause We therefore will identify \begin{center} \smath{\text{fn\ } a. X \;\approx\; \text{fn\ } b. \alert<3->{\swap{a}{b}}\act X} \end{center} provided that `\smath{b} is fresh for \smath{X} --- (\smath{b\fresh X})', i.e., does not occur freely in any ground term that might be substituted for \smath{X}.\bigskip\pause If we know more about \smath{X}, e.g., if we knew that \smath{a\fresh X} and \smath{b\fresh X}, then we can replace\\ \smath{\swap{a}{b}\act X} by \smath{X}. \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-4>[c] \frametitle{Equivalence Judgements} \alt<1>{Our equality is {\bf not} just}{but judgements} \begin{center} \begin{tabular}{rl} \colorbox{cream}{\smath{\onslide<2->{\nabla \vdash} t \approx t'}} & \alert{$\alpha$-equivalence}\\[1mm] \onslide<4->{\colorbox{cream}{\smath{\onslide<2->{\nabla \vdash} a \fresh t}}} & \onslide<4->{\alert{freshness}} \end{tabular} \end{center} \onslide<2->{ where \begin{center} \smath{\nabla = \{a_1\fresh X_1,\ldots, a_n\fresh X_n\}} \end{center} is a finite set of \alert{freshness assumptions}.} \onslide<3->{ \begin{center} \smath{\{a\fresh X,b\fresh X\} \vdash \text{fn\ } a. X \approx \text{fn\ } b. X} \end{center}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Rules for Equivalence} \begin{center} \begin{tabular}{c} Excerpt\\ (i.e.~only the interesting rules) \end{tabular} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Rules for Equivalence} \begin{center} \begin{tabular}{c} \colorbox{cream}{\smath{\infer{\nabla \vdash a \approx a}{}}}\\[8mm] \colorbox{cream}{% \smath{\infer{\nabla \vdash \abst{a}{t} \approx \abst{a}{t'}} {\nabla \vdash t \approx t'}}}\\[8mm] \colorbox{cream}{% \smath{\infer{\nabla \vdash \abst{a}{t} \approx \abst{b}{t'}} {a\not=b\;\; & \nabla \vdash t \approx \swap{a}{b}\act t'\;\;& \nabla \vdash a\fresh t'}}} \end{tabular} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3>[c] \frametitle{Rules for Equivalence} \begin{center} \colorbox{cream}{% \smath{% \infer{\nabla \vdash \pi\act X \approx \pi'\act X} {\begin{array}{c} (a\fresh X)\in\nabla\\ \text{for all}\; a \;\text{with}\;\pi\act a \not= \pi'\act a \end{array} }}} \end{center} \onslide<2->{ for example\\[4mm] \alt<2>{% \begin{center} \smath{\{a\fresh\!X, b\fresh\!X\} \vdash X \approx \swap{a}{b}\act X} \end{center}} {% \begin{center} \smath{\{a\fresh\!X, c\fresh\!X\} \vdash \swap{a}{c}\swap{a}{b}\act X \approx \swap{b}{c}\act X} \end{center}} \onslide<3->{ \begin{tabular}{@ {}lllll@ {}} because & \smath{\swap{a}{c}\swap{a}{b}}: & \smath{a\mapsto b} & \smath{\swap{b}{c}}: & \smath{a\mapsto a}\\ & & \smath{b\mapsto c} & & \smath{b\mapsto c}\\ & & \smath{c\mapsto a} & & \smath{c\mapsto b}\\ \end{tabular} disagree at \smath{a} and \smath{c}.} } \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Rules for Freshness} \begin{center} \begin{tabular}{c} Excerpt\\ (i.e.~only the interesting rules) \end{tabular} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{Rules for Freshness} \begin{center} \begin{tabular}{c} \colorbox{cream}{% \smath{\infer{\nabla \vdash a\fresh b}{a\not= b}}}\\[5mm] \colorbox{cream}{% \smath{\infer{\nabla \vdash a\fresh\abst{a}{t}}{}}}\hspace{7mm} \colorbox{cream}{% \smath{\infer{\nabla \vdash a\fresh\abst{b}{t}} {a\not= b\;\; & \nabla \vdash a\fresh t}}}\\[5mm] \colorbox{cream}{% \smath{\infer{\nabla \vdash a\fresh \pi\act X} {(\pi^{-1}\act a\fresh X)\in\nabla}}} \end{tabular} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-4>[t] \frametitle{$\approx$ is an Equivalence} \mbox{}\\[5mm] \begin{center} \colorbox{cream}{\alert{Theorem:} $\approx$ is an equivalence relation.} \end{center}\bigskip \only<1>{% \begin{tabular}{ll} (Reflexivity) & $\smath{\nabla\vdash t\approx t}$\\[2mm] (Symmetry) & if $\smath{\nabla\vdash t_1\approx t_2}\;$ then $\;\smath{\nabla\vdash t_2\approx t_1}$\\[2mm] (Transitivity) & if $\smath{\nabla\vdash t_1\approx t_2}\;$ and $\;\smath{\nabla\vdash t_2\approx t_3}$\\ & then $\smath{\nabla\vdash t_1\approx t_3}$\\ \end{tabular}} \only<2->{% \begin{itemize} \item<2-> \smath{\nabla \vdash t\approx t'} then \smath{\nabla \vdash \pi\act t\approx \pi\act t'} \item<2-> \smath{\nabla \vdash a\fresh t} then \smath{\nabla \vdash \pi\act a\fresh \pi\act t} \item<3-> \smath{\nabla \vdash t\approx \pi\act t'} then \smath{\nabla \vdash (\pi^{-1})\act t\approx t'} \item<3-> \smath{\nabla \vdash a\fresh \pi\act t} then \smath{\nabla \vdash (\pi^{-1})\act a\fresh t} \item<4-> \smath{\nabla \vdash a\fresh t} and \smath{\nabla \vdash t\approx t'} then \smath{\nabla \vdash a\fresh t'} \end{itemize} } \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-4> \frametitle{Comparison $=_\alpha$} Traditionally \smath{=_\alpha} is defined as \begin{center} \colorbox{cream}{% \begin{minipage}{9cm} \raggedright least congruence which identifies \smath{\abst{a}{t}} with \smath{\abst{b}{[a:=b]t}} provided \smath{b} is not free in \smath{t} \end{minipage}} \end{center} where \smath{[a:=b]t} replaces all free occurrences of\\ \smath{a} by \smath{b} in \smath{t}. \bigskip \only<2>{% \begin{textblock}{13}(1.2,10) For \alert{ground} terms: \begin{center} \colorbox{cream}{% \begin{minipage}{9.0cm} \begin{tabular}{@ {}rl} \underline{Theorem:} & \smath{t=_\alpha t'\;\;} if\hspace{-0.5mm}f~\smath{\;\;\emptyset \vdash t\approx t'}\\[2mm] & \smath{a\not\in F\hspace{-0.9mm}A(t)\;\;} if\hspace{-0.5mm}f~\smath{\;\;\emptyset\vdash a\fresh t} \end{tabular} \end{minipage}} \end{center} \end{textblock}} \only<3>{% \begin{textblock}{13}(1.2,10) In general \smath{=_\alpha} and \smath{\approx} are distinct! \begin{center} \colorbox{cream}{% \begin{minipage}{6.0cm} \smath{\abst{a}{X}=_\alpha \abst{b}{X}\;} but not\\[2mm] \smath{\emptyset \vdash \abst{a}{X} \approx \abst{b}{X}\;} (\smath{a\not=b}) \end{minipage}} \end{center} \end{textblock}} \only<4>{ \begin{textblock}{6}(1,2) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{10cm}\raggedright That is a crucial point: if we had\\[-2mm] \[\smath{\emptyset \vdash \abst{a}{X}\approx \abst{b}{X}}\mbox{,}\] then applying $\smath{[X:=a]}$, $\smath{[X:=b]}$, $\ldots$\\ give two terms that are {\bf not} $\alpha$-equivalent.\\[3mm] The freshness constraints $\smath{a\fresh X}$ and $\smath{b\fresh X}$ rule out the problematic substitutions. Therefore \[\smath{\{a\fresh X,b\fresh X\} \vdash \abst{a}{X}\approx \abst{b}{X}}\] does hold. \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-9> \frametitle{Substitution} \begin{tabular}{l@ {\hspace{8mm}}r@ {\hspace{1.5mm}}c@ {\hspace{1.5mm}}l@ {}} \pgfuseshading{smallbluesphere} & \smath{\sigma(\abst{a}{t})} & \smath{\dn} & \smath{\abst{a}{\sigma(t)}}\\[2mm] \pgfuseshading{smallbluesphere} & \smath{\sigma(\pi\act X)} & \smath{\dn} & \smath{\begin{cases}% \pi\;\act\;\sigma(X) & \!\!\text{if\ } \sigma(X)\not=X\\ \pi\act X & \!\!\text{otherwise}% \end{cases}}\\[6mm] \end{tabular}\bigskip\bigskip \pause \only<2-5>{ \only<2->{for example} \def\arraystretch{1.3} \begin{tabular}{@ {\hspace{14mm}}l@ {\hspace{3mm}}l} \onslide<2->{\textcolor{white}{$\Rightarrow$}} & \onslide<2->{\alt<3>{\smath{\underline{\abst{a}{\swap{a}{b}\act X}\;\,[X:=\pair{b}{Y}]}}} {\smath{\abst{a}{\swap{a}{b}\act X}\;\,[X:=\pair{b}{Y}]}}}\\ \onslide<3->{\smath{\Rightarrow}} & \onslide<3->{\alt<3,4>{\smath{\abst{a}{\underline{\swap{a}{b}\act X[X:=\pair{b}{Y}]}}}} {\smath{\abst{a}{\swap{a}{b}\act X}[X:=\pair{b}{Y}]}}}\\ \onslide<4->{\smath{\Rightarrow}} & \onslide<4->{\alt<4>{\smath{\abst{a}{\swap{a}{b}\act \underline{\pair{b}{Y}}}}} {\smath{\abst{a}{\underline{\swap{a}{b}}\act \pair{b}{Y}}}}}\\ \onslide<5->{\smath{\Rightarrow}} & \onslide<5->{\smath{\abst{a}{\pair{a}{\swap{a}{b}\act Y}}}} \end{tabular}} \only<6-> {\begin{tabular}{l@ {\hspace{8mm}}l@ {}} \pgfuseshading{smallbluesphere} & if \smath{\nabla\vdash t\approx t'} and\hspace{-2mm}\mbox{} \raisebox{-2.7mm}{ \alt<7>{\begin{tikzpicture} \draw (0,0) node[inner sep=1mm,fill=cream, very thick, draw=red, rounded corners=3mm] {\smath{\;\nabla'\vdash\sigma(\nabla)\;}}; \end{tikzpicture}} {\begin{tikzpicture} \draw (0,0) node[inner sep=1mm,fill=white, very thick, draw=white, rounded corners=3mm] {\smath{\;\nabla'\vdash\sigma(\nabla)\;}}; \end{tikzpicture}}}\\ & then \smath{\nabla'\vdash\sigma(t)\approx\sigma(t')} \end{tabular}} \only<9> {\begin{tabular}{l@ {\hspace{8mm}}l@ {}} \\[-4mm] \pgfuseshading{smallbluesphere} & \smath{\sigma(\pi\act t)=\pi\act\sigma(t)} \end{tabular}} \only<7>{ \begin{textblock}{6}(10,10.5) \begin{tikzpicture} \draw (0,0) node[inner sep=1mm,fill=cream, very thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{3.8cm}\raggedright this means\\[1mm] \smath{\nabla'\vdash a\fresh\sigma(X)}\\[1mm] holds for all\\[1mm] \smath{(a\fresh X)\in\nabla} \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Equational Problems} An equational problem \[ \colorbox{cream}{\smath{t \eqprob t'}} \] is \alert{solved} by \begin{center} \begin{tabular}{ll} \pgfuseshading{smallbluesphere} & a substitution \smath{\sigma} (terms for variables)\\[3mm] \pgfuseshading{smallbluesphere} & {\bf and} a set of freshness assumptions \smath{\nabla} \end{tabular} \end{center} so that \smath{\nabla\vdash \sigma(t)\approx \sigma(t')}. \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> Unifying equations may entail solving \alert{freshness problems}. \bigskip E.g.~assuming that \smath{a\not=a'}, then \[ \smath{\abst{a}{t}\eqprob \abst{a'}{t'}} \] can only be solved if \[ \smath{t\eqprob \swap{a}{a'}\act t'} \quad\text{\emph{and}}\quad \smath{a\freshprob t'} \] can be solved. \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Freshness Problems} A freshness problem \[ \colorbox{cream}{\smath{a \freshprob t}} \] is \alert{solved} by \begin{center} \begin{tabular}{ll} \pgfuseshading{smallbluesphere} & a substitution \smath{\sigma}\\[3mm] \pgfuseshading{smallbluesphere} & and a set of freshness assumptions \smath{\nabla} \end{tabular} \end{center} so that \smath{\nabla\vdash a \fresh \sigma(t)}. \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3> \frametitle{Existence of MGUs} \underline{Theorem}: There is an algorithm which, given a nominal unification problem \smath{P}, decides whether\\ or not it has a solution \smath{(\sigma,\nabla)}, and returns a \\ \alert{most general} one if it does.\bigskip\bigskip \only<3>{ Proof: one can reduce all the equations to `solved form' first (creating a substitution), and then solve the freshness problems (easy).} \only<2>{ \begin{textblock}{6}(2.5,9.5) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{8cm}\raggedright \alert{most general:}\\ straightforward definition\\ ``if\hspace{-0.5mm}f there exists a \smath{\tau} such that \ldots'' \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1> \frametitle{Remember the Quiz?} \textcolor{gray}{Assuming that $a$ and $b$ are distinct variables,\\ is it possible to find $\lambda$-terms $M_1$ to $M_7$ that make the following pairs $\alpha$-equivalent?} \begin{tabular}{@ {\hspace{14mm}}p{12cm}} \begin{itemize} \item \smath{\lambda a.\lambda b. (M_1\,b)\;} and \smath{\lambda b.\lambda a. (a\,M_1)\;} \item \textcolor{gray}{$\lambda a.\lambda b. (M_2\,b)\;$ and $\lambda b.\lambda a. (a\,M_3)\;$} \item \textcolor{gray}{$\lambda a.\lambda b. (b\,M_4)\;$ and $\lambda b.\lambda a. (a\,M_5)\;$} \item \smath{\lambda a.\lambda b. (b\,M_6)\;} and \smath{\lambda a.\lambda a. (a\,M_7)\;} \end{itemize} \end{tabular} \textcolor{gray}{If there is one solution for a pair, can you describe all its solutions?} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Answers to the Quiz} \small \def\arraystretch{1.6} \begin{tabular}{c@ {\hspace{2mm}}l} & \only<1>{\smath{\lambda a.\lambda b. (M_1\,b)\;} and \smath{\;\lambda b.\lambda a. (a\,M_1)}}% \only<2->{\smath{\abst{a}{\abst{b}{\pair{M_1}{b}}} \;\eqprob\; \abst{b}{\abst{a}{\pair{a}{M_1}}}}}\\ \onslide<3->{\smath{\redu{\id}}} & \only<3>{\smath{\abst{b}{\pair{M_1}{b}} \eqprob \alert{\swap{a}{b}} \act \abst{a}{\pair{a}{M_1}}\;,\;a\freshprob \abst{a}{\pair{a}{M_1}}}}% \only<4->{\smath{\abst{b}{\pair{M_1}{b}} \eqprob \abst{b}{\pair{b}{\swap{a}{b}\act M_1}}\;,\ a\freshprob \abst{a}{\pair{a}{M_1}}}}\\ \onslide<5->{\smath{\redu{\id}}} & \only<5->{\smath{\pair{M_1}{b} \eqprob \pair{b}{\swap{a}{b}\act M_1}\;,\;% a\freshprob \abst{a}{\pair{a}{M_1}}}}\\ \onslide<6->{\smath{\redu{\id}}} & \only<6->{\smath{M_1 \eqprob b \;,\; b \eqprob \swap{a}{b}\act M_1\;,\;% a\freshprob \abst{a}{\pair{a}{M_1}}}}\\ \onslide<7->{\smath{\redu{[M_1:=b]}}} & \only<7>{\smath{b \eqprob \swap{a}{b}\act \alert{b}\;,\;% a\freshprob \abst{a}{\pair{a}{\alert{b}}}}}% \only<8->{\smath{b \eqprob a\;,\; a\freshprob \abst{a}{\pair{a}{b}}}}\\ \onslide<9->{\smath{\redu{}}} & \only<9->{\smath{F\hspace{-0.5mm}AIL}} \end{tabular} \only<10>{ \begin{textblock}{6}(2,11) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{9cm}\raggedright \smath{\lambda a.\lambda b. (M_1\,b)} \smath{=_\alpha} \smath{\lambda b.\lambda a. (a\,M_1)} has no solution \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Answers to the Quiz} \small \def\arraystretch{1.6} \begin{tabular}{c@ {\hspace{2mm}}l} & \only<1>{\smath{\lambda a.\lambda b. (b\,M_6)\;} and \smath{\;\lambda a.\lambda a. (a\,M_7)}}% \only<2->{\smath{\abst{a}{\abst{b}{\pair{b}{M_6}}} \;\eqprob\; \abst{a}{\abst{a}{\pair{a}{M_7}}}}}\\ \onslide<3->{\smath{\redu{\id}}} & \only<3->{\smath{\abst{b}{\pair{b}{M_6}} \eqprob \abst{a}{\pair{a}{M_7}}}}\\ \onslide<4->{\smath{\redu{\id}}} & \only<4->{\smath{\pair{b}{M_6} \eqprob \pair{b}{\swap{b}{a}\act M_7}\;,\;b\freshprob\pair{a}{M_7}}}\\ \onslide<5->{\smath{\redu{\id}}} & \only<5->{\smath{b\eqprob b\;,\; M_6 \eqprob \swap{b}{a}\act M_7\;,\;% b\freshprob \pair{a}{M_7}}}\\ \onslide<6->{\smath{\redu{\id}}} & \only<6->{\smath{M_6 \eqprob \swap{b}{a}\act M_7\;,\;% b\freshprob \pair{a}{M_7}}}\\ \onslide<7->{\makebox[0mm]{\smath{\redu{[M_6:=\swap{b}{a}\act M_7]}}}} & \only<7->{\smath{\qquad b\freshprob \pair{a}{M_7}}}\\ \onslide<8->{\smath{\redu{\varnothing}}} & \only<8->{\smath{b\freshprob a\;,\;b\freshprob M_7}}\\ \onslide<9->{\smath{\redu{\varnothing}}} & \only<9->{\smath{b\freshprob M_7}}\\ \onslide<10->{\makebox[0mm]{\smath{\redu{\{b\fresh M_7\}}}}} & \only<10->{\smath{\;\;\varnothing}}\\ \end{tabular} \only<10>{ \begin{textblock}{6}(6,9) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{7cm}\raggedright \smath{\lambda a.\lambda b. (b\,M_6)\;} \smath{=_\alpha} \smath{\;\lambda a.\lambda a. (a\,M_7)}\\[2mm] we can take \smath{M_7} to be any $\lambda$-term that does not contain free occurrences of \smath{b}, so long as we take \smath{M_6} to be the result of swapping all occurrences of \smath{b} and \smath{a} throughout \smath{M_7} \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Properties} \begin{itemize} \item An interesting feature of nominal unification is that it does not need to create new atoms.\bigskip \begin{center}\small \colorbox{cream}{ \smath{\{a.t \eqprob b.t'\}\cup P \redu{\id} \{t \eqprob \swap{a}{b}\act t', a \freshprob t'\} \cup P}} \end{center}\bigskip\bigskip \pause \item The alternative rule \begin{center}\small \colorbox{cream}{ \begin{tabular}{@ {}l@ {}} \smath{\{a.t \eqprob b.t'\}\cup P \redu{\id}}\\ \mbox{}\hspace{2cm}\smath{\{\swap{a}{c}\act t \eqprob \swap{b}{c}\act t', c \freshprob t, c \freshprob t'\} \cup P} \end{tabular}} \end{center} leads to a more complicated notion of mgu.\medskip\pause \footnotesize \smath{\{a.X \eqprob b.Y\} \redu{} (\{a\fresh Y, c\fresh Y\}, [X:=\swap{a}{c}\swap{b}{c}\act Y])} \end{itemize} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3> \frametitle{Is it Useful?} Yes. $\alpha$Prolog by James Cheney (main developer)\bigskip\bigskip \color{darkgray} \begin{tabular}{@ {}l} type (Gamma, var(X), T) :- member (X,T) Gamma.\smallskip\medskip\\ type (Gamma, app(M, N), T') :-\\ \hspace{3cm}type (Gamma, M, arrow(T, T')),\\ \hspace{3cm}type (Gamma, N, T).\smallskip\medskip\\ type (Gamma, lam(\alert{x.M}), arrow(T, T')) / \alert{x \# Gamma} :-\\ \hspace{3cm}type ((x, T)::Gamma, M, T').\smallskip\medskip\\ member X X::Tail.\\ member X Y::Tail :- member X Tail.\\ \end{tabular} \only<2->{ \begin{textblock}{6}(1.5,0.5) \begin{tikzpicture} \draw (0,0) node[inner sep=3mm,fill=cream, ultra thick, draw=red, rounded corners=2mm] {\color{darkgray} \begin{minipage}{9cm}\raggedright {\bf One problem:} If we ask whether \begin{center} ?- type ([(x, T')], lam(x.Var(x)), T) \end{center} is typable, we expect an answer for T.\bigskip \onslide<3>{Solution: Before back-chaining freshen all variables and atoms in a program (clause).} \end{minipage}}; \end{tikzpicture} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Equivariant Unification} James Cheney proposed \begin{center} \colorbox{cream}{ \smath{t \eqprob t' \redu{\nabla, \sigma, \pi} \nabla \vdash \sigma(t) \approx \pi \act \sigma(t')}} \end{center}\bigskip\bigskip \pause But he also showed this problem is undecidable\\ in general. :( \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-> \frametitle{Taking Atoms as Variables} Instead of \smath{a.X}, have \smath{A.X}.\bigskip \pause Unfortunately this breaks the mgu-property: \begin{center} \smath{a.Z \eqprob X.Y.v(a)} \end{center} can be solved by \begin{center} \smath{[X:=a, Z:=Y.v(a)]} and \smath{[Y:=a, Z:=Y.v(Y)]} \end{center} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{HOPU vs. NOMU} \begin{itemize} \item James Cheney showed\bigskip \begin{center} \colorbox{cream}{\smath{HOPU \Rightarrow NOMU}} \end{center}\bigskip \item Jordi Levy and Mateu Villaret established\bigskip \begin{center} \colorbox{cream}{\smath{HOPU \Leftarrow NOMU}} \end{center}\bigskip \end{itemize} The translations `explode' the problems quadratically. \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1> \small\tt \begin{minipage}{13cm} \begin{tabular}{@ {\hspace{-2mm}}p{11.5cm}} \\ From: Zhenyu Qian <zhqian@microsoft.com>\\ To: Christian Urban <urbanc@in.tum.de>\\ Subject: RE: Linear Higher-Order Pattern Unification\\ Date: Mon, 14 Apr 2008 09:56:47 +0800\\ \\ Hi Christian,\\ \\ Thanks for your interests and asking. I know that that paper is complex. As I told Tobias when we met last time, I have raised the question to myself many times whether the proof could have some flaws, and so making it through a theorem prover would definitely bring piece to my mind (no matter what the result would be). The only problem for me is the time.\\ \ldots\\ Thanks/Zhenyu \end{tabular} \end{minipage} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1> \frametitle{Complexity} \begin{itemize} \item Christiopher Calves and Maribel Fernandez showed first that it is polynomial and then also quadratic \item Jordi Levy and Mateu Villaret showed that it is quadratic by a translation into a subset of NOMU and using ideas from Martelli/Montenari. \end{itemize} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1->[c] \frametitle{Conclusion} \begin{itemize} \item Nominal Unification is a completely first-order language, but implements unification modulo $\alpha$. \textcolor{gray}{(verification\ldots Ramana Kumar and Michael Norrish)} \medskip\pause \item NOMU has been applied in term-rewriting and logic programming. \textcolor{gray}{(Maribel Fernandez et al has a KB-completion procedure.)} I hope it will also be used in typing systems.\medskip\pause \item NOMU and HOPU are `equivalent' (it took a long time and considerable research to find this out).\medskip\pause \item The question about complexity is still an ongoing story.\medskip \end{itemize} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1>[c] \frametitle{ \begin{tabular}{c} \mbox{}\\[23mm] \alert{\LARGE Thank you very much!}\\ \alert{\Large Questions?} \end{tabular}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}text_raw {* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \mode<presentation>{ \begin{frame}<1-3> \frametitle{Most General Unifiers} \underline{Definition}: For a unification problem \smath{P}, a solution \smath{(\sigma_1,\nabla_1)} is \alert{more general} than another solution \smath{(\sigma_2,\nabla_2)}, iff~there exists a substitution \smath{\tau} with \begin{center} \begin{tabular}{ll} \pgfuseshading{smallbluesphere} & \alt<2>{\smath{\alert{\nabla_2\vdash\tau(\nabla_1)}}} {\smath{\nabla_2\vdash\tau(\nabla_1)}}\\ \pgfuseshading{smallbluesphere} & \alt<3>{\smath{\alert{\nabla_2\vdash\sigma_2\approx \tau\circ\sigma_1}}} {\smath{\nabla_2\vdash\sigma_2\approx \tau\circ\sigma_1}} \end{tabular} \end{center} \only<2>{ \begin{textblock}{13}(1.5,10.5) \smath{\nabla_2\vdash a\fresh \sigma(X)} holds for all \smath{(a\fresh X)\in\nabla_1} \end{textblock}} \only<3>{ \begin{textblock}{11}(1.5,10.5) \smath{\nabla_2\vdash \sigma_2(X)\approx \sigma(\sigma_1(X))} holds for all \smath{X\in\text{dom}(\sigma_2)\cup\text{dom}(\sigma\circ\sigma_1)} \end{textblock}} \end{frame}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% *}(*<*)end(*>*)