some changes

inputs/lecture_11.tex

@@ -173,10 +173,8 @@ i.e.~we want to associate a tree $T \subseteq  \N^{<\N}$}%
 Let \vocab{$\Tr$} $ \coloneqq  \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
 
 \begin{observe}
-    \[
-    \Tr \subseteq  {2^{\N}}^{<\N}
-    \]
-    is closed (where we take the topology of the Cantor space).
+    $\Tr \subseteq  {2^{\N}}^{<\N}$ is closed
+    (where we take the topology of the Cantor space).
 \end{observe}
 \gist{%
 Indeed, for any $ s \in \N^{<\N}$

inputs/lecture_13.tex

@@ -33,7 +33,6 @@ with $(f^{-1}(\{1\}), <)$.
     and  $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
 \end{proof}
 
-% TODO ANKI-MARKER
 
 \begin{definition}[\vocab{Kleene-Brouwer ordering}]
     Let $(A,<)$ be a linear order and $A$ countable.
@@ -58,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
     $(T, <_{KB}\defon{T})$ is well ordered.
 \end{proposition}
 \begin{proof}
-    If $T$ is ill-founded and $x \in [T]$,
-    then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
-    Thus $(T, <_{KB}\defon{T})$ is not well ordered.
+    \gist{%
+        If $T$ is ill-founded and $x \in [T]$,
+        then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
+        Thus $(T, <_{KB}\defon{T})$ is not well ordered.
 
-    Conversely, let $<\defon{KB}$ be not a well-ordering
-    on $T$.
-    Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
-    be an infinite descending chain.
-    We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
-    stabilizes for $n > n_0$.
-    Let $a_0 \coloneqq  s_{n_0}(0)$.
-    Now for $n \ge  n_0$ we have that $s_n(0)$ is constant,
-    hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
-    Thus there is $n_1 \ge n_0$ such that $s_n(1)$
-    is constant for all $n \ge n_1$.
-    Let $a_1 \coloneqq s_{n_1}(1)$
-    and so on.
-    Then $(a_0,a_1,a_2, \ldots) \in [T]$.
+        Conversely, let $<\defon{KB}$ be not a well-ordering
+        on $T$.
+        Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
+        be an infinite descending chain.
+        We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
+        stabilizes for $n > n_0$.
+        Let $a_0 \coloneqq  s_{n_0}(0)$.
+        Now for $n \ge  n_0$ we have that $s_n(0)$ is constant,
+        hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
+        Thus there is $n_1 \ge n_0$ such that $s_n(1)$
+        is constant for all $n \ge n_1$.
+        Let $a_1 \coloneqq s_{n_1}(1)$
+        and so on.
+        Then $(a_0,a_1,a_2, \ldots) \in [T]$.
+    }{easy}
 \end{proof}
+
+% TODO ANKI-MARKER
+
 \begin{theorem}[Lusin-Sierpinski]
     The set $\LO \setminus \WO$
     (resp.~$2^{\Q} \setminus \WO$)

jrpie-gist.sty

@@ -4,6 +4,7 @@
 % TODO gist info
 % TODO link to long version (provide link to main document)
 
+% TODO \phantomsection to cross link
 \newcommand{\gist}[2]{%
     \ifcsname EnableGist\endcsname%
         #2%