Problem_Set_01_Solutions

Problem_Set_01_Solutions - %LaTeX...

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%LaTeX \documentclass[12pt]{amsart} \usepackage{fullpage} \usepackage{amsmath} \newcommand{\tab}{\makebox[4em]{}} \newcommand{\ov}[1]{\overline{#1}} \newcommand{\ra}{\rightarrow} \newcommand{\lra}{\leftrightarrow} \ % ---------------- % PARAGRAPH INDENTATION \setlength{\parindent}{0in} \ % SPACE BETWEEN PARAGRAPHS \setlength{\parskip}{\medskipamount} \ % ---------------------------------------------------------------- % LISP CODE DISPLAYS. % Lisp code displays are enclosed between \bid and \eid. % Most characters are taken verbatim, in typewriter font, % Except: % Commands are still available (beginning with \) % Math mode is still available (beginning with $) % \outer\def\beginlisp{% \begin{minipage}[t]{\linewidth} \begin{list}{$\bullet$}{% \setlength{\topsep}{0in} \setlength{\partopsep}{0in} \setlength{\itemsep}{0in} \setlength{\parsep}{0in} \setlength{\leftmargin}{1.5em} \setlength{\rightmargin}{0in} \setlength{\itemindent}{0in} }\item[] \obeyspaces \obeylines \footnotesize\tt} \outer\def\endlisp{% \end{list} \end{minipage} } {\obeyspaces\gdef {\ }} { \begin{document} \begin{center} \textbf{ArsDigita University} \ \textbf{Month 2: Discrete Mathematics - Professor Shai Simonson} \ \textbf{Problem Set 1 -- ANSWERS} \ By: Dimitri Kountourogiannis B \end{center}
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\ \begin{enumerate} \item { Logic Proofs } \begin{enumerate} \item Prove that $a \rightarrow b$ is equivalent to $\neg b \rightarrow \neg a$ using a truth table. \\ \noindent {\bf ANSWER:} We compute the truth table values and observe they are the same. Note that $\neg b \rightarrow \neg a$ is called the contrapositive of $a \rightarrow b$. (Check out \\ http://logik.phl.univie.ac.at/$\sim$chris/formular-uk- zentral.html \\ for an automated truth-table generator) \[ \begin{array}{c|c|c|c|c|c} T & T & F & F & T & T\\ T & F & F & T & F & F\\ F & T & T & F & T & T\\ F & F & T & T & T & T \end{array} \] \item Prove it using {\em algebraic} identities. \\ \noindent {\bf ANSWER:} Using the equivalence rules on pages 17-18 of Rosen, \begin{align*} \end{align*} \item Prove that $a \rightarrow b$ is not equivalent to $ b \rightarrow a$.\\ \noindent {\bf ANSWER:} This is the sort of thing that is easy to show with a truth table (semantically) and is extremely difficult to show using the algebraic identities (syntactically). If $a \rightarrow b$ were equivalent to $b \rightarrow a$, then the two would have to agree for all possible {\em true/false} assignments of $a$ and $b$. But this is not the case: if we let $a$ be {\em true} and $b$ be {\em false}, then $a \rightarrow b$ evaluates to {\em false}, and $b \rightarrow a$ evaluates to {\em true}. So the two formulas are not equivalent.
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Problem_Set_01_Solutions - %LaTeX...

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