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-rw-r--r--library/set/cantor.tex2
-rw-r--r--library/set/regularity.tex2
2 files changed, 1 insertions, 3 deletions
diff --git a/library/set/cantor.tex b/library/set/cantor.tex
index 5455e70..093adee 100644
--- a/library/set/cantor.tex
+++ b/library/set/cantor.tex
@@ -16,7 +16,5 @@
Take $a'$ such that $a'\in A$ and $f(a') = B$ by \cref{surj}.
We have if $a'\in B$, then $a'\notin B$ by assumption.
We have if $a'\notin B$, then $a'\in B$ by assumption.
- We have $a'\in B$ by assumption.
- We have $a'\notin B$ by assumption.
Contradiction by assumption.
\end{proof}
diff --git a/library/set/regularity.tex b/library/set/regularity.tex
index 440467d..907bf57 100644
--- a/library/set/regularity.tex
+++ b/library/set/regularity.tex
@@ -28,7 +28,7 @@
\end{proof}
-% Isabelle/ZF-style foundation for case analysis
+% Foundation for case analysis
\begin{theorem}[Foundation]\label{foundation}
Let $A$ be a set.
Then $A = \emptyset$ or there exists $a\in A$