Add quantified calcs and relax tokenization

Use an `\iff`-calc to speed up `union_as_unions`. Also remove `in_implies_neq` which seems to interact badly with the choice axiom used by superposition-based proofs like Vampire in cut-down problems.

3 files changed, 10 insertions, 29 deletions

diff --git a/library/ordinal.tex b/library/ordinal.tex
index c092fa8..f978257 100644
--- a/library/ordinal.tex
+++ b/library/ordinal.tex
@@ -235,13 +235,6 @@ To show that \in\ is a strict total order it only remains to show that \in\ is c
         % Goal: ordinal(fgamma)=>(elem(falpha,fgamma)|elem(fgamma,falpha)|falpha=fgamma))
         %
         Assume $\gamma$ is an ordinal.
-        % Goal:
-        % elem(falpha,fgamma)|elem(fgamma,falpha)|falpha=fgamma
-        %
-        % Original Vampire proof:
-        % Follows by \cref{setext,transitiveset,ordinal,in_implies_neq,prec_is_ordinal,in_asymmetric}.
-        %
-        % Pruned proof:
         Follows by \cref{setext,transitiveset,ordinal}.
     \end{subproof}
 \end{proof}
@@ -489,7 +482,7 @@ Then $\alpha\subseteq\beta$.
 \end{proposition}
 \begin{proof}
     % Vampire proof:
-    Follows by \cref{inters_of_ordinals_is_ordinal,in_implies_neq,inters_iff_forall,ordinal_subseteq_implies_elem_or_eq,inters_subseteq_elem}.
+    Follows by \cref{inters_of_ordinals_is_ordinal,in_irrefl,inters_iff_forall,ordinal_subseteq_implies_elem_or_eq,inters_subseteq_elem}.
 \end{proof}
 
 \begin{proposition}\label{inters_of_ordinals_is_minimal}
diff --git a/library/set.tex b/library/set.tex
index 1f7a76a..d964ce7 100644
--- a/library/set.tex
+++ b/library/set.tex
@@ -124,16 +124,6 @@ which applies it to goals of the form “$A = B$” and  “$A \neq B$”.
     Suppose $A\neq\emptyset$. Then $A$ is inhabited.
 \end{proposition}
 
-%\begin{proposition}\label{empty_iff_emptyset}
-%    $A$ is empty iff $A = \emptyset$.
-%\end{proposition}
-
-
-\begin{proposition}%
-\label{empty_eq}
-    If $x$ and $y$ are empty, then $x = y$.
-\end{proposition}
-
 \begin{proposition}\label{emptyset_subseteq}
     For all $a$ we have $\emptyset \subseteq a$.
 \end{proposition}
@@ -351,17 +341,19 @@ The $\operatorname{\textsf{cons}}$ operation is determined by the following axio
     If $c\in B$, then $c\in A\union B$.
 \end{proposition}
 
-% TODO SLOW
 \begin{proposition}\label{union_as_unions}
     $\unions{\{x,y\}} = x\union y$.
 \end{proposition}
 \begin{proof}
-    %For all z we have
-    %\begin{align*}
-    %    z\in \unions{\{x,y\}}
-    %    &\iff \exists Z\in\{x,y\}. z\in Z
-    %\end{align*}
-    Follows by set extensionality.
+    For all $z$ we have
+    \begin{align*}
+        z\in \unions{\{x,y\}}
+        &\iff \exists Z\in\{x,y\}. z\in Z
+        &\iff z\in x \lor z\in y
+        &\iff z\in x\union y.
+    \end{align*}
+    Follows by \cref{setext}.
+    %Follows by set extensionality.
 \end{proof}
 
 \begin{proposition}[Commutativity of union]%
diff --git a/library/set/regularity.tex b/library/set/regularity.tex
index 068d6a8..10f4a1f 100644
--- a/library/set/regularity.tex
+++ b/library/set/regularity.tex
@@ -59,10 +59,6 @@
     Straightforward.
 \end{proof}
 
-\begin{proposition}\label{in_implies_neq}
-    If $a\in A$, then $a\neq A$.
-\end{proposition}
-
 \begin{proposition}\label{in_asymmetric}
     For all sets $a, b$ such that $a\in b$ we have $b\notin a$.
 \end{proposition}