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strategy: wip
[topbloke-formulae.git]
/
simple.tex
diff --git
a/simple.tex
b/simple.tex
index ae6fc5551fd797dedeeeced6423cf5708b0d0f05..b22fdbe9d15bc65321f58b6cd85ae65fb3ba250f 100644
(file)
--- a/
simple.tex
+++ b/
simple.tex
@@
-56,7
+56,6
@@
$$
\end{cases}
$$
\proofstarts
\end{cases}
$$
\proofstarts
-~
Firstly, if $L \haspatch \p$, $\exists_{F \in \py} F \le L$
and this $F$ is also $\le C$
Firstly, if $L \haspatch \p$, $\exists_{F \in \py} F \le L$
and this $F$ is also $\le C$
@@
-79,17
+78,22
@@
Ancestors: $ D \le C \equiv D \le L $.
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
+OK.
\subsubsection{For $L \nothaspatch \p$:}
Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
\subsubsection{For $L \nothaspatch \p$:}
Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
-Now by
contents of $L$, $D \notin L$, so $D \not
in C$.
+Now by
$\nothaspatch$, $D \not\isin L$ so $D \not\is
in C$.
OK.
$\qed$
OK.
$\qed$
+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies. $\qed$
+
\subsection{Foreign Inclusion:}
Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Inclusion:}
Simple Foreign Inclusion applies. $\qed$