diff --git a/chapter_1/1_3/definition_1_3_4.md b/chapter_1/1_3/definition_1_3_4.md new file mode 100644 index 0000000..424d24c --- /dev/null +++ b/chapter_1/1_3/definition_1_3_4.md @@ -0,0 +1,69 @@ +# 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. diff --git a/chapter_1/1_3/discrete_math_de_morgans_law_equivalency.png b/chapter_1/1_3/discrete_math_de_morgans_law_equivalency.png new file mode 100644 index 0000000..0311768 Binary files /dev/null and b/chapter_1/1_3/discrete_math_de_morgans_law_equivalency.png differ diff --git a/chapter_1/1_3/discrete_math_double_negation.png b/chapter_1/1_3/discrete_math_double_negation.png new file mode 100644 index 0000000..947d7fa Binary files /dev/null and b/chapter_1/1_3/discrete_math_double_negation.png differ diff --git a/chapter_1/1_3/discrete_math_implications_are_injunctions.png b/chapter_1/1_3/discrete_math_implications_are_injunctions.png new file mode 100644 index 0000000..2065d90 Binary files /dev/null and b/chapter_1/1_3/discrete_math_implications_are_injunctions.png differ diff --git a/chapter_1/1_3/discrete_math_negation_and_implication.png b/chapter_1/1_3/discrete_math_negation_and_implication.png new file mode 100644 index 0000000..8beb513 Binary files /dev/null and b/chapter_1/1_3/discrete_math_negation_and_implication.png differ diff --git a/chapter_1/1_3/discrete_math_quantifiers_and_negation.png b/chapter_1/1_3/discrete_math_quantifiers_and_negation.png new file mode 100644 index 0000000..74296bd Binary files /dev/null and b/chapter_1/1_3/discrete_math_quantifiers_and_negation.png differ diff --git a/chapter_1/1_3/double_negation.md b/chapter_1/1_3/double_negation.md new file mode 100644 index 0000000..719da11 --- /dev/null +++ b/chapter_1/1_3/double_negation.md @@ -0,0 +1,5 @@ +# Double Negation + +$$ \neg\neg P \equiv P $$ + +Example: "It is not the case that $c$ is not odd" means $c is odd.$ diff --git a/chapter_1/1_3/implications_are_disjunctions.md b/chapter_1/1_3/implications_are_disjunctions.md new file mode 100644 index 0000000..3d28eaa --- /dev/null +++ b/chapter_1/1_3/implications_are_disjunctions.md @@ -0,0 +1,8 @@ +# 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." diff --git a/chapter_1/1_3/investigate.md b/chapter_1/1_3/investigate.md new file mode 100644 index 0000000..029b879 --- /dev/null +++ b/chapter_1/1_3/investigate.md @@ -0,0 +1,102 @@ +# 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. diff --git a/chapter_1/1_3/preview_activity.md b/chapter_1/1_3/preview_activity.md new file mode 100644 index 0000000..a0b0f3d --- /dev/null +++ b/chapter_1/1_3/preview_activity.md @@ -0,0 +1,168 @@ +# 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. diff --git a/chapter_1/1_3/quantifiers_and_negation.md b/chapter_1/1_3/quantifiers_and_negation.md new file mode 100644 index 0000000..930cbc5 --- /dev/null +++ b/chapter_1/1_3/quantifiers_and_negation.md @@ -0,0 +1,5 @@ +# Quantifiers and Negation. + +$\neg \forall x P(x)$ is equivalent to $\exists x \neg P(x)$. + +$\neg \exists x P(x)$ is equivalent to $\forall x \neg P(x)$. diff --git a/chapter_1/1_3/theorem_1_3_7.md b/chapter_1/1_3/theorem_1_3_7.md new file mode 100644 index 0000000..38ad060 --- /dev/null +++ b/chapter_1/1_3/theorem_1_3_7.md @@ -0,0 +1,10 @@ +# 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,_ + +$$ \neg(P \vee Q) \equiv \neg P \wedge \neg Q $$ diff --git a/chapter_1/1_3/theorem_1_3_9.md b/chapter_1/1_3/theorem_1_3_9.md new file mode 100644 index 0000000..66e9ab9 --- /dev/null +++ b/chapter_1/1_3/theorem_1_3_9.md @@ -0,0 +1,8 @@ +# 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._ diff --git a/leftoff.txt b/leftoff.txt index 900731f..987e7ca 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -64 +77