150 lines
5.2 KiB
TeX
150 lines
5.2 KiB
TeX
% File: quiz3.tex
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% Created: 15:57:17 Sat, 18 Sep 2021 EDT
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% Last Change: 15:57:17 Sat, 18 Sep 2021 EDT
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%
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\documentclass[letterpaper]{article}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{cancel}
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\usepackage{amssymb}
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\usepackage{listings}
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\usepackage[shortlabels]{enumitem}
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\usepackage{lipsum}
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\usepackage{soul}
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\usepackage{geometry}
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\geometry{portrait, margin=1in}
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\date{09/18/2021}
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\title{%
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Quiz \#3\\
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\large PHIL-205-01:Symbolic Logic}
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\author{Blizzard MacDougall}
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\begin{document}
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\maketitle
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\pagenumbering{arabic}
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\section{Section 1: Translations in Truth Functional Logic}
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Using the following symbolization key, please symbolize the following sentences.
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$B$: Your shifters aren't compatible with 12 speeds in the back.
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$C$: Campagnolo doesn't offer a 52T cog.
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$F$: Bikes really are "machines for freedom".
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$M$: You need a monster cog in the back.
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$O$: You ought to install a 1x drivetrain on your bike.
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$S$: You're better off with a Shimano drivetrain.
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$V$: You can find a vintage SunTour derailleur on eBay.
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\begin{enumerate}
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\item You ought to install a 1x drivetrain on your bike only if you need a monster cog in the back.\\
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\begin{equation}
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O\implies M
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\end{equation}
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\item If you need a monster cog in the back but Compagnolo doesn't offer a 52T cog, then you're better off with a Shimano drivetrain.\\
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\begin{equation}
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(M\land C)\implies S
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\end{equation}
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\item Either your shifters aren't compatible with 12 speeds in the back and you ought to install a 1x drivetrain on your bike, or Campagnolo doesn't offer a 52T cog and you're better off with a Shimano drivetrain.\\
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\begin{equation}
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(B\land O)\lor(C\land S)
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\end{equation}
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\item You ought to install a 1x drivetrain on your bike if either Campagnolo doesn't offer a 52T cog or your shifters aren't compatible with 12 speeds in the back (but not both).\\
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\begin{equation}
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(\neg[C\iff B])\implies O
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\end{equation}
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\item If a) you're better off with a Shimano drivetrain only if your shifters aren't compatible with 12 speeds in the back, and b) you ought to install a 1x drivetrain on your bike unless you cannot find a vintage SunTour derailleur on eBay, then bikes really are "machines of freedom" only if you don't need a monster cog in the back.
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\begin{equation}
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([B\implies S]\land[O\lor V])\implies(M\implies F)
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\end{equation}
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\end{enumerate}
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\newpage
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\section{Section 2: Translation in First Order Logic}
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Using the following symbolization key symbolize the following sentences.
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Domain: Songs
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$Cx$: x was written by Tracy Chapman
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$Sx$: x is, without a doubt, one of the best songs of the 1980s
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$Bx$: x was recently covered by Black Pumas
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$f$: "Fast Car"
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\begin{enumerate}
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\item "Fast Car" is, without a doubt, one of the best songs of the 1980s if and only if it was recently covered by Black Pumas.\\
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\begin{equation}
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Sf\iff Bf
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\end{equation}
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\item Tracy Chapman wrote a song that is, without a doubt, one of the best songs of the 1980s.\\
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\begin{equation}
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\exists x(Cx\land Sx)
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\end{equation}
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\item Black Pumas recently covered a Tracy Chapman song that is, without a doubt, one of the best songs of the 1980s.\\
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\begin{equation}
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\exists x(Cx\land Sx\land Bx)
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\end{equation}
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\item Every song recently covered by Black Pumas was written by Tracy Chapman.\\
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\begin{equation}
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\forall x(Bx\implies Cx)
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\end{equation}
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\item Not every song recently covered by Black Pumas was written by Tracy Chapman.\\
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\begin{equation}
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\neg(\forall x[Bx\implies Cx])
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\end{equation}
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\item No song recently covered by Black Pumas was written by Tracy Chapman.\\
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\begin{equation}
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\neg(\exists x[Bx\land Cx])
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\end{equation}
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\item If Black Pumas recently covered a song that is, without a doubt, one of the best songs of the 1980s, then "Fast Car" is, without a doubt, one of the best songs of the 1980s.\\
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\begin{equation}
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\exists x(Bx\land Sx)\implies Sf
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\end{equation}
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\item Either "Fast Car" was not recently covered by Black Pumas, or there's a song that is, without a doubt, neither one of the best songs of 1980s not written by Tracy Chapman.\\
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\begin{equation}
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Bf\lor(\exists x[(\neg Sx)\land(\neg Cx)])
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\end{equation}
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\end{enumerate}
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\newpage
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\section{Section 3: Truth Table}
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Is the following a valid argument?
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\begin{equation}
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\begin{split}
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P\land R\\
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\neg R\lor Q\\
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\therefore P\implies Q
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\end{split}
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\end{equation}
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\begin{center}
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\begin{tabular}{|c|c|c||c|c|c||c|c|c||c|c|c|}
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\hline
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$P$ & $Q$ & $R$ & \multicolumn{3}{c||}{$P\land R$} & \multicolumn{3}{c||}{$\neg R\lor Q$} & \multicolumn{3}{c|}{$P\implies Q$}\\
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\hline
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1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\
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\hline
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1 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
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\hline
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1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
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\hline
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1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0\\
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\hline
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0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1\\
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\hline
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0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1\\
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\hline
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0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\
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\hline
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0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0\\
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\hline
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\end{tabular}
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\end{center}
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There are two lines in the truth table where the conclusion is false. In both of these lines, the premises are not both true. Therefore, this argument is valid.
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\end{document}
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