Visually highlighting symbolic material

Question

I have a diagram in which I need to indicate the difference between symbolic mathematics and English words (potentially to someone who doesn't know about it):

Colour works beautifully for making this distinction. Unfortunately, for the version that is going to go in a book, I can't use colour and have been told not to use greyscale (which was my first fallback). I've tried bold, but it looks pretty horrible and doesn't highlight the distinction terribly well:

(Also bold is bad because sometimes it has mathematical significance.)

Underlining just looks really horrible:

I would be very grateful for any alternative suggestions on how to visually highlight symbolic material.

Source in case it's useful:

\documentclass{article}
\usepackage{xcolor}
\usepackage{framed}
\begin{document}

\definecolor{textual-colour}{rgb}{0,0,0}
\definecolor{symbolic-colour}{rgb}{0.2,0.2,0.9}
\newcommand{\marktextual}[1]{{\color{textual-colour}#1}}
\newcommand{\marksymbolic}[1]{{\color{symbolic-colour}#1}}

\begin{framed}
\marktextual{{If}  \marksymbolic{$K \leq G$} {and} {there} {are} {inclusions} \marksymbolic{$gKg^{-1}\leq K$}  {for} {every} \marksymbolic{$g\in G$}, {then} \marksymbolic{$K \triangleleft G$}: {replacing} \marksymbolic{$g$} {by} \marksymbolic{$g^{-1}$}, {we} {have} {the} {inclusion} \marksymbolic{$g^{-1}Kg\leq K$}, {and} {this} {gives} {the} {reverse} {inclusion} \marksymbolic{$K\leq gKg^{-1}$}.}
\smallskip

\marktextual{{The} {kernel} \marksymbolic{$K$} {of} {a} {homomorphism} \marksymbolic{$f:G\rightarrow H$} {is} {a} {normal}  {subgroup}: {if} \marksymbolic{$a\in K$}, {then} \marksymbolic{$f(a)=1$}; {if} \marksymbolic{$g\in G$}, {then} \marksymbolic{$f(gag^{-1}) = f(g)f(a)f(g^{-1}) =  f(g)f(g^{-1}) = 1$}, {and} {so} \marksymbolic{$gag^{-1}\in K$}. {Hence}, \marksymbolic{$gKg^{-1}\leq K$}  {for} {all} \marksymbolic{$g\in G$}, {and} {so} \marksymbolic{$K \triangleleft G$}. {Conversely}, {we} {shall} {see} {later} {that} {every} {normal} {subgroup} {is} {the} {kernel} {of} {some} {homomorphism}.}
\end{framed}\end{document}

Edit: while I was trying to keep this short, from the comments it looks like a little more explanation of the context is necessary. The text of the book says:

"At first sight, the most striking feature of mathematical language is the way in which it mixes material that looks as if it is drawn from a natural language with material built up out of idiosyncratically mathematical symbols. The distinction is illustrated in Figure 2.1."

Figure 2.1 looks like this:

Answer 1

A solution with hf-tikz (requires two compilation runs):

\documentclass[a4paper]{article}  
\usepackage{framed}
\usepackage[customcolors]{hf-tikz}
\hfsetbordercolor{white}
\definecolor{vlgray}{rgb}{0.87 0.87 0.87}
\usepackage[colorlinks=true,urlcolor=blue]{hyperref}

\newcounter{highlight}
\newcommand{\highlight}[1]{%
\stepcounter{highlight}\tikzmarkin{\thehighlight}(0.04,-0.1)(-0.04,0.3)#1\tikzmarkend{\thehighlight}%
}

\newcommand{\exampletext}{%
  If \highlight{$K \leq G$} and there are inclusions \highlight{$gKg^{-1}\leq K$} 
  for every \highlight{$g\in G$}, then \highlight{$K \triangleleft G$}: replacing 
  \highlight{$g$} by \highlight{$g^{-1}$}, we have the inclusion 
  \highlight{$g^{-1}Kg\leq K$}, and this gives the reverse inclusion 
  \highlight{$K\leq gKg^{-1}$}.

  The kernel \highlight{$K$} of a homomorphism \highlight{$f:G\rightarrow H$} is a 
  normal  subgroup: if \highlight{$a\in K$}, then \highlight{$f(a)=1$}; if 
  \highlight{$g\in G$}, then 
  \highlight{$f(gag^{-1}) = f(g)f(a)f(g^{-1}) =$}\penalty\relpenalty
  \highlight{$ f(g)f(g^{-1}) = 1$},
  and so \highlight{$gag^{-1}\in K$}. Hence, \highlight{$gKg^{-1}\leq K$} for all 
  \highlight{$g\in G$}, and so \highlight{$K \triangleleft G$}. Conversely, 
  we shall see later that every normal subgroup is the kernel of some homomorphism.}

\begin{document}
\hfsetfillcolor{white}
\begin{framed}
\exampletext
\end{framed}

\hfsetfillcolor{vlgray}
\begin{framed}
\exampletext
\end{framed}

Applied to \href{http://tex.stackexchange.com/questions/88977/putting-a-coloured-background-behind-text-without-adding-whitespace}{Putting a coloured background behind text without adding whitespace}:

Lorem ipsum \highlight{dolor} sit amet, \highlight{consectetuer} adipiscing elit, sed diam 

Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam

\end{document}

The result:

Answer 2

Without using color for the formulas or a gray background, I don't see any good way for distinguishing between text and formulas. Boxing the formulas seems out of the question because it's really too heavy; underlining is, as you say, horrible.

A way out might be to present the same paragraphs typeset twice: one with blanks in place of the formulas, so by comparing the two images one can see what you're referring to.

\documentclass{article}
\usepackage{framed,calc}

\newcommand{\exampletext}{%
  If \marksymbolic{$K \leq G$} and there are inclusions \marksymbolic{$gKg^{-1}\leq K$} 
  for every \marksymbolic{$g\in G$}, then \marksymbolic{$K \triangleleft G$}: replacing 
  \marksymbolic{$g$} by \marksymbolic{$g^{-1}$}, we have the inclusion 
  \marksymbolic{$g^{-1}Kg\leq K$}, and this gives the reverse inclusion 
  \marksymbolic{$K\leq gKg^{-1}$}.

  The kernel \marksymbolic{$K$} of a homomorphism \marksymbolic{$f:G\rightarrow H$} is a 
  normal  subgroup: if \marksymbolic{$a\in K$}, then \marksymbolic{$f(a)=1$}; if 
  \marksymbolic{$g\in G$}, then 
  \marksymbolic{$f(gag^{-1}) = f(g)f(a)f(g^{-1}) =$}\penalty\relpenalty
  \marksymbolic{$ f(g)f(g^{-1}) = 1$},
  and so \marksymbolic{$gag^{-1}\in K$}. Hence, \marksymbolic{$gKg^{-1}\leq K$} for all 
  \marksymbolic{$g\in G$}, and so \marksymbolic{$K \triangleleft G$}. Conversely, 
  we shall see later that every normal subgroup is the kernel of some homomorphism.}

\begin{document}
\begin{framed}
\providecommand{\marksymbolic}[1]{\makebox[\widthof{#1}]{\hrulefill}}
\exampletext
\end{framed}

\begin{framed}
\providecommand{\marksymbolic}[1]{\mbox{#1}}
\exampletext
\end{framed}

\end{document}

Here is the same with

\providecommand{\marksymbolic}[1]{{\fboxsep=0pt\colorbox[gray]{0.9}{#1}}}

Answer 3

The advise of "not to use greyscale" probably can be extended to the background, and personally I think so many gray boxes breaks the reading flow.

My proposal is use a clearly different fonts for the math and text modes, if needed, adjusting the size of math mode to look similar to text mode. Although one might think a higher size would highlight also the math mode, may be is only less readable. For example, for me with lxfonts look better a smaller font size:

\documentclass{article}
\usepackage{lxfonts} 
\usepackage{framed}
\newcommand{\marksymbolic}[1]{{\footnotesize #1}}
\begin{document}
\usefont{T1}{lmr}{m}{n}
\begin{framed}
If 
\marksymbolic{$K \leq G$} and there are inclusions 
\marksymbolic{$gKg^{-1}\leq K$} for every 
\marksymbolic{$g\in G$}, then 
\marksymbolic{$K \triangleleft G$}: replacing 
\marksymbolic{$g$} by 
\marksymbolic{$g^{-1}$}, we have the inclusion 
\marksymbolic{$g^{-1}Kg\leq K$}, and this gives the reverse inclusion
\marksymbolic{$K\leq gKg^{-1}$}.

The kernel 
\marksymbolic{$K$} of a homomorphism 
\marksymbolic{$f:G\rightarrow H$} is a normal subgroup: if 
\marksymbolic{$a\in K$}, then 
\marksymbolic{$f(a)=1$}; if 
\marksymbolic{$g\in G$}, then 
\marksymbolic{$f(gag^{-1}) = f(g)f(a)f(g^{-1}) =  f(g)f(g^{-1}) = 1$}, and so 
\marksymbolic{$gag^{-1}\in K$}. Hence, 
\marksymbolic{$gKg^{-1}\leq K$}  for all 
\marksymbolic{$g\in G$}, and so 
\marksymbolic{$K \triangleleft G$}. 
Conversely, we shall see later that every normal subgroup is the kernel of some homomorphism.
\end{framed}
\end{document}