grammar: visually separate syntax and meta-syntax

grammar/grammar.tex

@@ -9,6 +9,19 @@
 \date{}
 
 \renewcommand{\grammarlabel}[2]{#1 #2}
+\usepackage{xspace}
+
+\newcommand{\stx}[1]{\texttt{#1}} % For concrete syntax
+\newcommand{\meta}[1]{\emph{#1}} % For concrete syntax
+
+\def\e/{\meta{e}}
+\def\a/{\meta{a}}
+\def\b/{\meta{b}}
+\def\c/{\meta{c}}
+\def\p/{\meta{p}}
+\def\q/{\meta{q}}
+\def\f/{\meta{f}}
+\def\x/{\meta{x}}
 
 \begin{document}
 
@@ -16,35 +29,35 @@
 
 \begin{grammar}
 <e> ::=
-    x | c
-  \alt e.a | e.a or e
-  \alt p:e | e e
-  \alt [ e ... e ] | \{ e = e; ...; e = e; \}
-  \alt with e; e
-  \alt if e then e else e
-  \alt let x = e; ...; x = e; in e
-  \textcolor{blue}{\alt e :: e | nil}
-  \alt attrNames e | listToAttrs e
-  \alt addAttr e e e | removeAttr e e
+    \x/ | \c/
+  \alt \e/\stx{.}\a/ | \e/\stx{.}\a/ \stx{or} \e/
+  \alt \p/\stx{:}\e/ | \e/ \e/
+  \alt \stx{[} \e/ \ldots{} \e/ \stx{]} | \stx{\{} \e/ \stx{=} \e/\stx{;} ...\stx{;} \e/ \stx{=} \e/\stx{; \}}
+  \alt \stx{with} \e/\stx{;} \e/
+  \alt \stx{if} \e/ \stx{then} \e/ \stx{else} \e/
+  \alt \stx{let} \x/ \stx{=} \e/\stx{; \ldots{};} \x/ \stx{=} \e/\stx{; in} \e/
+  \textcolor{blue}{\alt \e/ \stx{::} \e/ | \stx{nil}}
+  \alt \stx{attrNames} \e/ | \stx{listToAttrs} \e/
+  \alt \stx{addAttr} \e/ \e/ \e/ | \stx{removeAttr} \e/ \e/
   % Not as-if in nix, but simpler than the actual builtins ("//" or
   % removeAttrs), and yet sufficient thanks to attrNames and listToAttrs
-  \alt deepSeq e e
-  \alt functionArgs e
+  \alt \stx{deepSeq} \e/ \e/
+  \alt \stx{functionArgs} \e/
   % Maybe this can be typed as a function, but it is rather unlikely
 
-<a> ::= b. ... .b
+<a> ::= \b/\stx{.} \ldots{} \stx{.}\b/
 
-<b> ::= \$\{e\} | l
+<b> ::= \stx{\$\{}\e/\stx{\}} | \meta{l}
 
-<c> ::= s | i | ...
-  \alt typeOf
-  \alt ...
+<c> ::= \meta{s} | \meta{i} | \ldots{}
+  \alt \stx{typeOf}
+  \alt \ldots{}
 
-<p> ::= q | q@x | x
+<p> ::= \q/ | \q/\stx{@}\x/ | \x/
 
-<q> ::= \{ f, ..., f \} | \{ f, ..., f, … \}
+<q> ::= \stx{\{} \f/\stx{, \ldots{},} \f/ \stx{\}} | \stx{\{} \f/\stx{, \ldots{},} \f/ \stx{, ...\}}
 
-<f> ::= x | x ? e
+<f> ::= \x/ | \x/ \stx{?} \e/
 
 \end{grammar}
 \end{document}