merge remove redundant element in condition

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 3da2db93f9e42a580cbe201f3889de5d0037cda5..5459b88726b9e2264e74bd0261cd67bc54ae34b3 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -496,8 +496,7 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
    \begin{cases}
       R \in \py : & \baseof{R} \ge \baseof{L}
               \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
-      R \in \pn : & R \ge \baseof{L}
-              \land M = \baseof{L} \\
+      R \in \pn : & M = \baseof{L} \\
       \text{otherwise} : & \false
    \end{cases}
 }\]
@@ -552,7 +551,7 @@ That is, $\baseof{C} = \baseof{R}$.
 
 \subsubsection{For $R \in \pn$:}
 
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
 $A \le \baseof{L} \implies A \le R$, so
 $A \le R \lor A \le \baseof{L} \equiv A \le R$.  
 Thus $A \le C \equiv A \le R$.