\overline{B} -> \overline{A}

Typo in Intermediate Logic, Section 4.9

Author: FnControlOption

content/sets-functions-relations/size-of-sets/comparing-size.tex

@@ -115,12 +115,12 @@
   is always some number that is !!a{element} of one of $Z_n$ or
   $\overline{Z}$ but not the other. We follow the same idea here,
   except the indices~$n$ are now !!{element}s of~$A$ instead
-  of~$\PosInt$. The set $\overline{B}$ is defined so that it is
+  of~$\PosInt$. The set $\overline{A}$ is defined so that it is
   different from~$g(x)$ for each $x \in A$, because $x \in g(x)$ iff
-  $x \notin \overline{B}$. Again, there is always !!a{element} of~$A$
-  which is !!a{element} of one of $g(x)$ and $\overline{B}$ but not
+  $x \notin \overline{A}$. Again, there is always !!a{element} of~$A$
+  which is !!a{element} of one of $g(x)$ and $\overline{A}$ but not
   the other. And just as $\overline{Z}$ therefore cannot be on the
-  list $Z_1$, $Z_2$, \dots, $\overline{B}$ cannot be in the range
+  list $Z_1$, $Z_2$, \dots, $\overline{A}$ cannot be in the range
   of~$g$.}{}
 
 \oliflabeldef{sfr:siz:nen-alt:thm:nonenum-pownat}{It's instructive to