🚧 Going through 1.3

chapter_1/1_3/definition_1_3_4.md

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+# 1.3.3 Logical Equivalence
+
+You might have noticed in Example 1.3.2 that the final column in the truth table
+for $\neg P \vee Q$ is identical to the final column in the truth table for
+$P \to Q$:
+
+| $P$ | $Q$ | $P \to Q$ | $\neg P \vee Q$ |
+| --- | --- | --------- | --------------- |
+| T   | T   | T         | T               |
+| T   | F   | F         | F               |
+| F   | T   | T         | T               |
+| F   | F   | T         | T               |
+
+This says that no matter what $P$ and $Q$ are, the statements $\neg P \vee Q$
+and $P \to Q$ are either both true or both false. We therefore say these
+statements are **logically equivalent**.
+
+---
+
+## Definition 1.3.4 Logical Equivalence
+
+Two (molecular) statements $P$ and $Q$ are **logically equivalent** provided $P$
+is true precisely when $Q$ is true. That is, $P$ and $Q$ have the same truth
+value under any assignment of truth values to their atomic parts. We write this
+as $P \equiv Q$.
+
+---
+
+To verify that two statements are logically equivalent, you can make a truth
+table for each and check whether the columns for the two statements are
+identical.
+
+In Section 1.2 we claimed that whenever an implication is true, so is its
+contrapositive. We can now make this claim as the following theorem.
+
+## Theorem 1.3.5
+
+_An implication is logically equivalent to its contrapositive. That is,_
+
+$$ P \to Q \equiv \neg Q \to \neg P $$
+
+**Proof**
+
+We simply examine the truth tables.
+
+| $P$ | $Q$ | $P \to Q$ |
+| --- | --- | --------- |
+| T   | T   | T         |
+| T   | F   | F         |
+| F   | T   | T         |
+| F   | F   | T         |
+
+| $P$ | $Q$ | $\neg Q$ | $\neg P$ | $\neg Q \to \neg P$ |
+| --- | --- | -------- | -------- | ------------------- |
+| T   | T   | F        | F        | T                   |
+| T   | F   | T        | F        | F                   |
+| F   | T   | F        | T        | T                   |
+| F   | F   | T        | T        | T                   |
+
+(Note that we have the truth value combinations in the same order in both
+tables, so we can easily see that the final columns are identical)
+
+(Note that we have the truth value combinations in the same order in both
+tables, so we can easily see that the final columns are identical)
+
+Recognizing two statements as logically equivalent can be quite helpful.
+Rephrasing a mathematical statement can often lend insight into what it is
+saying, or how to prove or refute it. By using truth tables we can
+systematically verify that two statements are indeed logically equivalent.

chapter_1/1_3/double_negation.md

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+# Double Negation
+
+$$ \neg\neg P \equiv P $$
+
+Example: "It is not the case that $c$ is not odd" means $c is odd.$

chapter_1/1_3/implications_are_disjunctions.md

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+# Implications are Disjunctions
+
+$$ P \to Q \equiv \neg P \vee Q $$
+
+.
+
+Example: "If a number is a multiple of 4, then it is even" is equivalent to, "A
+number is not a multiple of 4, or (else) it is even."

chapter_1/1_3/investigate.md

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+# Investigate!
+
+Holmes always wears one of the two vests he owns: one tweed and one mint green.
+He always wears either the green vest or red shoes. Whenever he wears a purple
+shirt and the green vest, he chooses to not wear a bow tie. He never wears the
+green vest unless he is also wearing either a purple shirt or red shoes.
+Whenever he wears red shoes, he also wears a purple shirt. Today, Holmes wore a
+bow tie. What else did he wear?
+
+## Try it 1.3.1
+
+Spend a few minutes thinking about the _Investigate!_ question above. Of the six
+statements in the puzzle, only one is atomic. Use this atomic statement and one
+other statement to deduce a new statement about what Holmes might (or might not)
+be wearing. Explain why you think your new statement is true.
+
+**Hint**
+
+The atomic statement is, "Holmes wore a bow tie." Only one of the molecular
+statement has this as one of its atoms.
+
+A:
+
+Let $B$ be "Holmes wears a bow tie." Also, let' $P$ be "Holmes wears a purple
+shirt" and $G$ be "Holmes wears a green vest".
+
+Based off the molecular statement:
+
+"Whenever he wears a purple shirt and the green vest, he chooses not to wear a
+bow tie."
+
+We can write this as:
+
+$$ (P \wedge G) \to \neg B $$
+
+But we know that $B$ is true from the problem statement, which means that Holmes
+is not wearing a purple shirt and the green vest:
+
+$$ \neg (P \wedge G) $$
+
+Let's now write out the other statements. Let $R$ be "Holmes wears the red
+shoes." We know from the problem statement that "He always wears either the
+green vest or red shoes." This is written as:
+
+$$ G \vee R $$
+
+Let $T$ be "Holmes wears the tweed vest." We know from the problem statement
+that "Holmes always wears one of the two vests he owns: one tweed and one mint
+green." This is written as:
+
+$$ T \vee G $$
+
+"He never wears the green vest unless he is also wearing either a purple shirt
+or red shoes.":
+
+$$ G \to (P \vee R) $$
+
+"Whenever he wears red shoes, he also wears a purple shirt."
+
+$$ R \to P $$
+
+This gives us everything we need, let's investigate what we know, and track back
+through the problem to find out what Holmes is wearing.
+
+$$ B \to \neg (P \wedge G) $$
+
+While Holmes could be wearing either the purple shirt or the green vest, he
+cannot wear them together. Let's assume he's wearing the green vest:
+
+$$ G \to (P \vee R) $$
+
+So he can't wear the purple shirt, but he can wear the red shoes.
+
+$$ R \to P $$
+
+Ah, that doesn't work, whenever Holmes wears the red shoes he also wears a
+purple shirt. Therefore Holmes cannot be wearing the green vest.
+
+$$ \neg G $$
+
+So, now we consider "He always wears either the green vest or red shoes."
+
+$$ G \vee R $$
+
+Since we know that Holmes isn't wearing the green vest, therefore he must be
+wearing the red shoes:
+
+$$ R $$
+
+And if he's wearing the red shoes, he is also wearing a purple shirt:
+
+$$ R \to P $$
+
+We also know that Holmes always either wears one of the two vests, the tweed or
+the mint green.
+
+$$ T \vee G $$
+
+Since we know he's not wearing the green vest, he must be wearing the tweed
+vest.
+
+So Holmes is wearing a tweed vest, a bow tie, a purple shirt, and red shoes.

chapter_1/1_3/preview_activity.md

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+# Preview Activity
+
+1.
+
+Q: Consider the statement, "Whenever Holmes wears a purple shirt and the green
+vest, he chooses to not wear a bow tie." Let $P$ be the statement, "Holmes wears
+a purple shirt," $G$ be the statement, "Holmes wears the green vest," and $B$ be
+the statement, "Holmes wears a bow tie." Which of the following is the best
+translation of the statement into propositional logic?
+
+A. $(P \wedge G) \to \neg B$
+
+B. $P \wedge G \to B$
+
+C. $(P \vee G) \to \neg B$
+
+D. $P \vee (G \to B)$
+
+A:
+
+A. $(P \wedge G) \to \neg B$
+
+This is the answer. $\wedge$ stands in for "and", and the given statement
+declares that whenever Holmes wears a purple shirt, $P$ and the green vest, $G$
+($P \wedge G$), then he chooses to not wear a bow to $\to \neg B$.
+
+B. $P \wedge G \to B$
+
+This does not translate properly, what this implies is that "Whenever Holmes
+wears a purple shirt and the green vest, he chooses to wear a bow tie." But the
+problem statement explicitly declares that Holmes does _not_ wear a bow tie if
+he is wearing a purple shirt and the green vest.
+
+C. $(P \vee G) \to \neg B$
+
+This also does not translate properly, this would be "Whenever Holmes wears a
+purple shirt or the green vest, he chooses to not wear a bow tie."
+
+D. $P \vee (G \to B)$
+
+This one's a bit more difficult to translate, but also is not the best
+translation. This essentially says "Either Holmes wears a purple shirt, or if
+Holmes wears a green vest, he wears a bow tie."
+
+2.
+
+Q: Consider the statement, "Holmes never wears the green vest unless he is also
+wearing either a purple shirt or red shoes." With $P$ and $G$ as in the previous
+question, and $R$ being the statement, "Holmes wears red shoes," which of the
+following is the best translation of the statement into propositional logic?
+
+A. $G \to (P \vee R)$
+
+B. $\neg G \to (P \vee R)$
+
+C. $(P \vee R) \to G$
+
+D. $(P \vee R) \to \neg G$
+
+A:
+
+A. $G \to (P \vee R)$
+
+This is equivalent to the given statement. This is saying "If Holmes is wearing
+the green vest, then he is either wearing a purple shirt or red shoes."
+
+B. $\neg G \to (P \vee R)$
+
+This is not the best translation, although the $\neg G$ makes it appear so. What
+this is saying is "Whenever Holmes is not wearing the green vest, then he is
+either wearing a purple shirt or red shoes."
+
+C. $(P \vee R) \to G$
+
+This appears to be the same as part A, but if we state this out carefully, this
+says "If Holmes wears either the purple shirt or the red shoes, then he is
+wearing the green vest." That is not the same as the given statement.
+
+D. $(P \vee R) \to \neg G$
+
+This is the same as part C, but it implies the opposite. "If Holmes is wearing
+either the purple shirt or the red shoes, then he is not wearing the green
+vest." Which is also not the same as the given statement.
+
+3.
+
+Q: Consider the statement, "If you major in math, then you will get a
+high-paying job," and the statement, "Either you don't major in math, or you
+will get a high-paying job." In which of the following cases are _both_
+statements true?
+
+A. You major in math and get a high-paying job.
+
+B. You major in math and don't get a high-paying job.
+
+C. You don't major in math and do get a high-paying job.
+
+D. You don't major in math and don't get a high-paying job.
+
+A :
+
+Let $P$ be "You major in math" and $Q$ be "You get a high-paying job."
+
+The first given statement then is:
+
+$$ P \to Q $$
+
+The second statement is:
+
+$$ \neg P \vee Q $$
+
+These are logically equivalent if you think about it.
+
+$P \to Q$ is false if $P$ is true and $Q$ is false.
+
+$\neg P \vee Q$ is false if $P$ is true and $Q$ is false.
+
+With this in mind, let's answer each section.
+
+A. You major in math and get a high-paying job.
+
+$P = \text{ T}$
+
+$Q = \text{ T}$
+
+$$ P \to Q = T \to T = T $$
+
+$$ \neg P \vee Q = F \vee T = T $$
+
+This is a case where both statements are true.
+
+B. You major in math and don't get a high-paying job.
+
+$P = \text{ T}$
+
+$Q = \text{ F}$
+
+$$ P \to Q = T \to F = F $$
+
+$$ \neg P \vee Q = F \vee F = F $$
+
+Both are false, so this is not a case where both statements are true..
+
+C. You don't major in math and do get a high-paying job.
+
+$P = \text{ F}$
+
+$Q = \text{ T}$
+
+$$ P \to Q = F \to T = T $$
+
+$$ \neg P \vee Q = T \vee T = T $$
+
+This is a case where both statements are true.
+
+D. You don't major in math and don't get a high-paying job.
+
+$P = \text{ F}$
+
+$Q = \text{ F}$
+
+$$ P \to Q = F \to F = T $$
+
+$$ \neg P \vee Q = T \vee F = T $$
+
+Both statements are true, this is a case.
+
+The answer is A, C, and D.

chapter_1/1_3/quantifiers_and_negation.md

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+# Quantifiers and Negation.
+
+$\neg \forall x P(x)$ is equivalent to $\exists x \neg P(x)$.
+
+$\neg \exists x P(x)$ is equivalent to $\forall x \neg P(x)$.

chapter_1/1_3/theorem_1_3_7.md

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+# Theorem 1.3.7 De Morgan's Laws.
+
+_The negation of a disjunction or conjunction is logically equivalent to a
+conjunction or disjunction of negations, respectively. That is,_
+
+$$ \neg(P \wedge Q) \equiv \neg P \vee \neg Q $$
+
+_and,_
+
+$$ \neg(P \vee Q) \equiv \neg P \wedge \neg Q $$

chapter_1/1_3/theorem_1_3_9.md

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+# Theorem 1.3.9 Negation of an Implication
+
+_The negation of an implication is a conjunction:_
+
+$$ \neg(P \to Q) \equiv P \wedge \neg Q $$
+
+_That is, the only way for an implication to be false is for the hypothesis to
+be true **AND** the conclusion to be false._

leftoff.txt

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-64
+77