\def\band{\bop{and}}
\def\bor{\bop{or}}
\def\bxor{\bop{xor}}
-\def\bnot{\bop{not}}
+\def\bnot{\mathop{\hbox{\sc not}}}
% A reversed version of \ddots with extra space at the top to get good alignment of exponents.
\def\rddots{\mathinner{\mkern1mu\raise\p@\vbox{\kern7\p@\hbox{.}}\mkern2mu
\n{$\<LSB>(x)$}{the position of the lowest bit set in~$x$ \[lsbmsb]}
\n{$\<MSB>(x)$}{the position of the highest bit set in~$x$ \[lsbmsb]}
\n{$\bf x$}{a~vector with elements $x_1,\ldots,x_d$; $x$ is its bitwise encoding \[vecnota]}
+\n{$\band$}{bitwise conjunction: $(x\band y)[i]=1$ iff $x[i]=1 \land y[i]=1$}
+\n{$\bor$}{bitwise disjunction: $(x\bor y)[i]=1$ iff $x[i]=1 \lor y[i]=1$}
+\n{$\bnot$}{bitwise negation: $(\bnot x)[i]=1-x[i]$}
+\n{$\bxor$}{bitwise non-equivalence: $(x\bxor y)[i]=1$ iff $x[i]\ne y[i]$}
+\n{$x \shl n$}{bitwise shift of~$x$ by $n$~positions to the left: $x\shl n = x\cdot 2^n$}
+\n{$x \shr n$}{bitwise shift of~$x$ by $n$~positions to the right: $x\shr n = \lfloor x/2^n \rfloor$}
}
%--------------------------------------------------------------------------------