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\noindent{Katie Burket}
\noindent{Foundations Homework 1: Problems 1.1, 2.3, and 2.5}
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\noindent \textbf{Problem 1.1}
\noindent \underline{Understanding the Problem}:
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We need to find a word that, when written in standard capital letters, satisfies two conditions: Its appearance needs to remain unchanged when it is reflected across a horizontal axis, and its appearance needs to remain unchanged when it is reflected across a vertical axis. The word needs to be found in a dictionary to be a valid solution.
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\noindent \underline{Devising a Plan}:
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In order for the word to satisfy both conditions, all the letters that make up the word must not change appearance when reflected horizontally or vertically. If any of the letters did change appearance, then the entire word would certainly not appear the same in all situations. Therefore, we need to go through the alphabet and pick out which letters appear the same when reflected vertically and horizontally. This would provide a list of letters that we know are the only letters that can make up the word. Also, in order for the word to satisfy the condition of not changing appearance when it is reflected across a vertical axis, we know that the word has to be a palindrome: that is, spelled the same foreward and backward. This is because the reflection across a vertical axis will make the last letter of the original word the first letter of the reflected word, and vice versa. If the word were not a palindrome, the reflected word would not spell out the same word as the original; it would be backward. From these two observations, we should be able to come up with a solution.
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\noindent \underline{Carrying out the Plan}:
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The first step is to come up with the list of letters that we can choose from to come up with a word. After examining all letters of the alphabet, these are the ones that appear the same when reflected both horizontally and vertically:
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H, I, O, X
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Therefore, the word we come up with must be made up of only letters from this list.
From here, we need to keep in mind that the word must be a palindrome as we brainstorm words. Possibly the best method for finding a word is to try as many combinations of letters as we can that seem like they could go together, and then decide if one of the combinations is actually a word and consult a dictionary to check. We know that most likely, if the word contains an H or an X, then those letters will need to have one or both of the vowels in between or around them. This is because a word made up of just the letters X and H does not likely exist and most words require vowels in between these consonants, especially X. So, the only combinations we can come up with involving the X or the H are:
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OHO, HOH, IHI, HIH, XOX, OXO, XIX, and IXI
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The amount of a certain letter in a row can be increased if we think that would create a word from one of these forms; for example, having the first combination include two of the letter H in a row instead of one: OHHO instead of OHO. However, regardless of adding extra letters or not, none of these words are necessarily recognizable without consulting a dictionary.
After consulting a dictionary, it turns out that OXO is in fact an English word. According to the Merriam-Webster dictionary, it is an adjective that means "containing oxygen." Therefore, our solution to this problem is:
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OXO
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This is a valid solution because it is a real word, when it is reflected across a horizontal axis it still appears as OXO, and when it is reflected across a vertical axis it still appears as OXO. Therefore, all conditions are met.
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\noindent \underline{Looking Back}:
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There are multiple solutions to this problem. We only considered the combinations of letters that seemed the most simple, since simple combinations are more likely to be real words. A simpler solution to the problem is the word: I. However, this seemed to be too straightforward and we wanted to try to come up with a better solution. Nonetheless, there are multiple solutions and this was just one way to reach one. We could have also tried to pick one of the letters to be the first (and at least last) letter in the word and only tried to form combinations for that letter, only trying a different first letter when we spent too much time on one and reached no solution. Many different approaches could have led us to an answer.
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\noindent \textbf{Problem 2.3}
The truth table for $\neg(P\wedge Q)$ appears as follows:
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\begin{tabular}{ | c | c | c | c | }
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$P$ & $Q$ & $P\wedge Q$ & $\neg(P\wedge Q)$\\
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$T$ & $T$ & $T$ & $F$\\
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$T$ & $F$ & $F$ & $T$\\
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$F$ & $T$ & $F$ & $T$\\
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$F$ & $F$ & $F$ & $T$\\
\hline
\end{tabular}
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The truth table for $\neg P\vee \neg Q$ appears as follows:
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\begin{tabular}{ | c | c | c | c | c | }
\hline
$P$ & $Q$ & $\neg P$ & $\neg Q$ & $\neg P \vee \neg Q$\\
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$T$ & $T$ & $F$ & $F$ & $F$\\
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$T$ & $F$ & $F$ & $T$ & $T$\\
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$F$ & $T$ & $T$ & $F$ & $T$\\
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$F$ & $F$ & $T$ & $T$ & $T$\\
\hline
\end{tabular}
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We can conclude that these two statement forms are equivalent because Theorem 2.7 tells us that they have the correct conditions to be equivalent. Another way to prove that these statement forms are equivalent is by realizing that the truth table for the equivalence of these two statement forms would have a final column all containing the letter T, which means $(\neg(P \wedge Q)) \leftrightarrow (\neg P \vee \neg Q)$ is a tautology. Any two statement forms that have an equivalence that is a tautology are equivalent statement forms. We know that $(\neg(P \wedge Q)) \leftrightarrow (\neg P \vee \neg Q)$ is a tautology because both statement forms have the same truth values in the final column, and for $P \leftrightarrow Q$ to be true, both P and Q need to have the same truth value. Therefore, the final column of the equivalence will be all T, and this show that $(\neg(P \wedge Q)) \leftrightarrow (\neg P \vee \neg Q)$ is a tautology and proves that these statement forms are equivalent.
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\noindent \textbf{Problem 2.5}
The truth table for $P \rightarrow \neg(Q \wedge \neg P)$ appears as follows:
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\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$P$ & $Q$ & $\neg P$ & $Q \wedge \neg P$ & $\neg(Q \wedge \neg P)$ & $P \rightarrow \neg(Q \wedge \neg P)$\\
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$T$ & $T$ & $F$ & $F$ & $T$ & $T$\\
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$T$ & $F$ & $F$ & $F$ & $T$ & $T$\\
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$F$ & $T$ & $T$ & $T$ & $F$ & $T$\\
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$F$ & $F$ & $T$ & $F$ & $T$ & $T$\\
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\end{tabular}
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Thus, the statement form $P \rightarrow \neg(Q \wedge \neg P)$ is a tautology because the final column consists of all the letter T.
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