From 10c15b2e9a42f164fac2d5ea67d37a99a9f53532 Mon Sep 17 00:00:00 2001
From: Akshay <nerdy@peppe.rs>
Date: Wed, 16 Dec 2020 10:28:32 +0530
Subject: wording

---
 presentation.tex | 7 +++----
 1 file changed, 3 insertions(+), 4 deletions(-)

diff --git a/presentation.tex b/presentation.tex
index 0ca6878..a6d3238 100644
--- a/presentation.tex
+++ b/presentation.tex
@@ -2,9 +2,9 @@
 
 \usepackage[utf8]{inputenc}
 
-
+\usefonttheme{serif}
 \title{Finite Automata and Formal Languages (18CS52)}
-\author{Akshay O}
+\author{Akshay O (1RV18CS016)}
 \institute[R V College of Engineering]
 {
     Submitted To: Dr. Krishnappa H K\\
@@ -24,13 +24,12 @@
 
 \begin{frame}
     \frametitle{Proof}
-    Let $M_1$ and $M_2$ be two minimal DFAs (have as few states as possible) accepting the language $L$. Both $M_1$ and $M_2$ must have exactly the same number of states because, if one had more states than the other, it would not be a minimal DFA.
+    Consider the union of $M_1$ and $M_2$ (assume there are no common names of states among the two). The transition function of the combined automaton is the union of the transition rules of $M_1$ and $M_2$. Run the state distinguishability process on this new DFA.
 \end{frame}
 
 \begin{frame}
     \frametitle{Proof}
 
-    For an empty input, $|w| = 0$, both DFAs remain in the start state.
     The start states of $M_1$ and $M_2$, $p_0$ and $q_0$ respectively; are indistinguishable because 
     \begin{equation}
         L(M_1) = L(M_2) = L
-- 
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