From 5a75e39799bb20bd4bc20a64bbf81ca3266ca05c Mon Sep 17 00:00:00 2001 From: tomit4 Date: Thu, 14 May 2026 18:44:24 -0700 Subject: [PATCH] :construction: Still going... --- chapter_1/1_3/additional_exercises.md | 498 ++++++++++++++++++++++++++ chapter_1/1_3/practice_problems.md | 275 ++++++++++++++ chapter_1/1_3/preview_activity.md | 4 +- chapter_1/1_3/reading_questions.md | 36 ++ leftoff.txt | 2 +- 5 files changed, 812 insertions(+), 3 deletions(-) create mode 100644 chapter_1/1_3/additional_exercises.md create mode 100644 chapter_1/1_3/practice_problems.md create mode 100644 chapter_1/1_3/reading_questions.md diff --git a/chapter_1/1_3/additional_exercises.md b/chapter_1/1_3/additional_exercises.md new file mode 100644 index 0000000..2696403 --- /dev/null +++ b/chapter_1/1_3/additional_exercises.md @@ -0,0 +1,498 @@ +# 1.3.8 Additional Exercises + +1. + +Q: You stumble upon two trolls playing Stratego. They tell you: + +Troll 1: If we are cousins, then we are both knaves. + +Troll 2: We are cousins, or we are both knaves. + +Could both trolls be knights? Recall that all trolls are either +always-truth-telling knights or always-lying knaves. Explain your answer and how +you can use truth tables to find it. + +Let $P$ be the predicate "We are cousins" and $Q$ be the predicate "We are both +knaves." + +Troll 1 then says: + +$$ P \to Q $$ + +Troll 2 says: + +$$ P \vee Q $$ + +We can use truth tables to actually explore all possibilities. + +| $P$ | $Q$ | $P \to Q$ | $P \vee Q$ | +| --- | --- | --------- | ---------- | +| T | T | T | T | +| T | F | F | T | +| F | T | T | T | +| F | F | T | F | + +In order for our tested conclusion, "Both trolls are knights" to be true, then +$Q$, "We are both knaves." must be false. + +| $P$ | $Q$ | $P \to Q$ | $P \vee Q$ | +| --- | --- | --------- | ---------- | +| T | F | F | T | +| F | F | T | F | + +Ah, but if we isolate for that here, we can see that at least one of our +premises is false, meaning that at least one of the trolls is lying, and +therefore it is not possible for both trolls to be knights. + +2. Next you come upon three trolls, helpfully wearing name tags. They say: + +Pat: "If either Quinn or I are knights, then so is Ryan." + +Quinn: "Ryan is a knight, and if Pat is a knight, then so am I." + +Ryan: "Quinn is a knave, but Pat and I share the same persuasion." + +Create a truth table that includes all three statements. Then use the truth +table to determine the persuasion of each troll. + +Let $P$ be "Pat is a knight", let "Q" be "Quinn is a knight", and let $R$ be +"Ryan is a knight." + +Pat claims: + +$$ (Q \vee P) \to R $$ + +Quinn claims: + +$$ R \wedge (P \to Q) $$ + +Ryan claims: + +$$ \neg Q \wedge (R \leftrightarrow P)$$ + +| $P$ | $Q$ | $R$ | $(Q \vee P)$ | $(P \to Q)$ | $\neg Q$ | $(R \leftrightarrow P)$ | $(Q \vee P) \to R$ | $R \wedge (P \to Q)$ | $\neg Q \wedge (R \leftrightarrow P)$ | +| --- | --- | --- | ------------ | ----------- | -------- | ----------------------- | ------------------ | -------------------- | ------------------------------------- | +| T | T | T | T | T | F | T | T | T | F | +| T | T | F | T | T | F | F | F | F | F | +| T | F | T | T | F | T | T | T | F | T | +| T | F | F | T | F | T | F | F | F | F | +| F | T | T | T | T | F | F | T | T | F | +| F | T | F | T | T | F | T | F | F | F | +| F | F | T | F | T | T | F | T | T | F | +| F | F | F | F | T | T | T | T | F | T | + +Now we check each Premise $P$, $Q$, and $R$ as True and compare against whether +their respective claims are also true: + +--- + +Pat: + +$$ (Q \vee P) \to R $$ + +| $P$ | $Q$ | $R$ | $(Q \vee P)$ | $(P \to Q)$ | $\neg Q$ | $(R \leftrightarrow P)$ | $(Q \vee P) \to R$ | $R \wedge (P \to Q)$ | $\neg Q \wedge (R \leftrightarrow P)$ | +| --- | --- | --- | ------------ | ----------- | -------- | ----------------------- | ------------------ | -------------------- | ------------------------------------- | +| T | T | T | T | T | F | T | T | T | F | +| T | F | T | T | F | T | T | T | F | T | + +In the first row: + +The assertion Pat makes, $P$, is true, and his claim $(Q \vee P) \to R$ is also +true. + +The assertion Quinn makes, $Q$, is true, and her claim $R \wedge (P \to R)$ is +true. + +The assertion Ryan makes, $R$, is true, and his claim +$\neg Q \wedge (R \leftrightarrow)$ is false. + +Because the validity of Ryan's assertion does not correspond with the validity +of his claim, this is not useful in determining the persuasion of each troll. + +In the second row: + +The assertion Pat makes, $P$, is true, and his claim $(Q \vee P) \to R$ is also +true. + +The assertion Quinn makes, $Q$, is false, and her claim $R \wedge (P \to R)$ is +false. + +The assertion Ryan makes, $R$, is true, and his claim +$\neg Q \wedge (R \leftrightarrow)$ is true. + +This last row demonstrates the persuasion of each troll. + +--- + +Quinn: + +$$ R \wedge (P \to Q) $$ + +| $P$ | $Q$ | $R$ | $(Q \vee P)$ | $(P \to Q)$ | $\neg Q$ | $(R \leftrightarrow P)$ | $(Q \vee P) \to R$ | $R \wedge (P \to Q)$ | $\neg Q \wedge (R \leftrightarrow P)$ | +| --- | --- | --- | ------------ | ----------- | -------- | ----------------------- | ------------------ | -------------------- | ------------------------------------- | +| T | T | T | T | T | F | T | T | T | F | +| F | T | T | T | T | F | F | T | T | F | + +In the first row: + +The assertion Pat makes, $P$, is true, and his claim $(Q \vee P) \to R$ is also +true. + +The assertion Quinn makes, $Q$, is true, and her claim $R \wedge (P \to R)$ is +true. + +The assertion Ryan makes, $R$, is true, and his claim +$\neg Q \wedge (R \leftrightarrow)$ is false. + +Because the validity of Ryan's assertion does not correspond with the validity +of his claim, this is not useful in determining the persuasion of each troll. + +In the second row: + +The assertion Pat makes, $P$, is false, and his claim $(Q \vee P) \to R$ is +true. + +Because the validity of Pat's assertion does not correspond with the validity of +his claim, this is not useful in determining the persuasion of each troll. + +--- + +Ryan: + +$$ \neg Q \wedge (R \leftrightarrow P)$$ + +| $P$ | $Q$ | $R$ | $(Q \vee P)$ | $(P \to Q)$ | $\neg Q$ | $(R \leftrightarrow P)$ | $(Q \vee P) \to R$ | $R \wedge (P \to Q)$ | $\neg Q \wedge (R \leftrightarrow P)$ | +| --- | --- | --- | ------------ | ----------- | -------- | ----------------------- | ------------------ | -------------------- | ------------------------------------- | +| T | F | T | T | F | T | T | T | F | T | + +The assertion Pat makes, $P$, is true, and corresponds with the claim Pat makes +$(Q \vee P) \to R$, and is true. This matches. + +The assertion that Quin makes, $Q$, is false, and corresponds with the claim +Quin makes $R \wedge (P \to Q)$. + +The assertion that Ryan makes, $R$, is true, and corresponds with the claim Ryan +makes $\neg Q \wedge (R \leftrightarrow P)$. + +This last row demonstrates the persuasion of each troll. + +In conclusion, we can confidently say that Pat is a knight, Quinn is a knave, +and Ryan is a knight. + +3. + +Q: Consider the statement about a party, "If it's your birthday or there will be +cake, then there will be cake." + +(a) Translate the above statement into symbols. Clearly state which statement is +$P$ and which is $Q$. + +(b) Make a truth table for the statement. + +\(c\) Assuming the statement is true, what (if anything) can you conclude if you +know there will be cake? + +(d) Assuming the statement is true, what (if anything) can you conclude if you +know there will not be cake? + +(e) Suppose you found out that the statement was a lie. What can you conclude? + +A: + +(a) Translate the above statement into symbols. Clearly state which statement is +$P$ and which is $Q$. + +Let $P$ be "It's your birthday", and let $Q$ be "There will be cake." + +So, in symbols, the given statement is: + +$$ P \vee Q \to Q $$ + +(b) Make a truth table for the statement. + +| $P$ | $Q$ | $P \vee Q$ | $P \vee Q \to Q$ | +| --- | --- | ---------- | ---------------- | +| T | T | T | T | +| T | F | T | F | +| F | T | T | T | +| F | F | F | T | + +\(c\) Assuming the statement is true, what (if anything) can you conclude if you +know there will be cake? + +| $P$ | $Q$ | $P \vee Q$ | $P \vee Q \to Q$ | +| --- | --- | ---------- | ---------------- | +| T | T | T | T | +| F | T | T | T | + +I can conclude that there will be cake whether or not it's my birthday. + +(d) Assuming the statement is true, what (if anything) can you conclude if you +know there will not be cake? + +| $P$ | $Q$ | $P \vee Q$ | $P \vee Q \to Q$ | +| --- | --- | ---------- | ---------------- | +| F | F | F | T | + +If I assume the statement $P \vee Q \to Q$ is true, and I know there will not be +cake $Q=F$, then I know it's not my birthday. + +(e) Suppose you found out that the statement was a lie. What can you conclude? + +| $P$ | $Q$ | $P \vee Q$ | $P \vee Q \to Q$ | +| --- | --- | ---------- | ---------------- | +| T | F | T | F | + +I can conclude that it is my birthday and that there is no cake. + +4. + +Q: Geoff Poshingten is out at a fancy pizza joint and decides to order a +calzone. When the waiter asks what he would like in it, he replies, "I want +either pepperoni or sausage. Also, if I have sausage, then I must also include +quail. Oh, and if I have pepperoni or quail, then I must also have ricotta +cheese." + +(a) Translate Geoff's order into logical symbols. + +(b) The waiter knows that Geoff is either a liar or a truth-teller (so either +everything he says is false, or everything is true). Which is it? + +\(c\) What, if anything, can the waiter conclude about the ingredients in +Geoff's desired calzone? + +A: + +(a) Translate Geoff's order into logical symbols. + +Let $P$ be "Geoff has pepperoni." Let $S$ be "Geoff has sausage." Let $Q$ be +"Geoff has quail." Let $R$ be "Geoff has Ricotta Cheese." + +"I want either pepperoni or sausage": + +$$ P \vee S $$ + +"If I have sausage, then I must also include quail." + +$$ S \to Q $$ + +"If I have pepperoni or quail, then I must also have ricotta cheese." + +$$ (P \vee Q) \to R $$ + +(b) The waiter knows that Geoff is either a liar or a truth-teller (so either +everything he says is false, or everything is true). Which is it? + +| $P$ | $S$ | $Q$ | $R | $P \vee S$ | $S \to Q$ | $P \vee Q$ | $(P \vee Q) \to R$ | +| --- | --- | --- | -- | ---------- | --------- | ---------- | ------------------ | +| T | T | T | T | T | T | T | T | +| T | T | T | F | T | T | T | F | +| T | T | F | T | T | F | T | T | +| T | T | F | F | T | F | T | F | +| T | F | T | T | T | T | T | T | +| T | F | T | F | T | T | T | F | +| T | F | F | T | T | T | T | T | +| T | F | F | F | T | T | T | F | +| F | T | T | T | T | T | T | T | +| F | T | T | F | T | T | T | F | +| F | T | F | T | T | F | F | T | +| F | T | F | F | T | F | F | T | +| F | F | T | T | F | T | T | T | +| F | F | T | F | F | T | T | F | +| F | F | F | T | F | T | F | T | +| F | F | F | F | F | T | F | T | + +Let's now filter where all the claims are true, $P \vee S$, $S \to Q$, and +$(P \vee Q) \to R$: + +| $P$ | $S$ | $Q$ | $R | $P \vee S$ | $S \to Q$ | $P \vee Q$ | $(P \vee Q) \to R$ | +| --- | --- | --- | -- | ---------- | --------- | ---------- | ------------------ | +| T | T | T | T | T | T | T | T | + +Let's now filter where all the claims are false, $P \vee S$, $S \to Q$, and +$(P \vee Q) \to R$: + +| $P$ | $S$ | $Q$ | $R | $P \vee S$ | $S \to Q$ | $P \vee Q$ | $(P \vee Q) \to R$ | +| --- | --- | --- | -- | ---------- | --------- | ---------- | ------------------ | +| | | | | | | | | + +And there are no rows where all three claims of Geoff's are F (i.e., he is +lying). So everything he says is true. Geoff is a truth-teller. + +\(c\) What, if anything, can the waiter conclude about the ingredients in +Geoff's desired calzone? + +Since we've concluded already that Geoff is a truth-teller, that means +everything he says is true, let's examine all possible claims where Geoff's +claims are true: + +So, breaking it down: + +$P \vee S$: Geoff is getting either Pepperoni or Sausage or both. + +$S \to Q$ If Geoff has Sausage, he also has Quail. + +$P \vee Q$: Geoff is getting either pepperoni or quail or both. + +$(P \vee Q) \to R$: If Geoff gets either pepperoni or Quail or both, he has +Ricotta cheese. + +| $P$ | $S$ | $Q$ | $R | $P \vee S$ | $S \to Q$ | $(P \vee Q) \to R$ | +| --- | --- | --- | -- | ---------- | --------- | ------------------ | +| T | T | T | T | T | T | T | +| T | F | T | T | T | T | T | +| T | F | F | T | T | T | T | +| F | T | T | T | T | T | T | + +The only thing that is consistent is that Geoff gets Ricotta cheese. We are +unable to determine which other ingredients that Geoff will include in his +Calzone. + +5. + +Q: Determine whether the following two statements are logically equivalent: +$\neg (P \to Q)$ and $P \wedge \neg Q$. Explain how you know you are correct. + +A: + +We can use a truth table: + +| $P$ | $Q$ | $(P \to Q)$ | $\neg (P \to Q)$ | $\neg Q$ | $P \wedge \neg Q$ | +| --- | --- | ----------- | ---------------- | -------- | ----------------- | +| T | T | T | F | F | F | +| T | F | F | T | T | T | +| F | T | T | F | F | F | +| F | F | T | F | T | F | + +And then we just compare the two statement's rows: + +| $\neg (P \to Q)$ | $P \wedge \neg Q$ | +| ---------------- | ----------------- | +| F | F | +| T | T | +| F | F | +| F | F | + +This is also the same problem presented in Example 1.3.8, which asks us to prove +logical equivalency without truth tables. + +Let's first take our first statement: + +$$ \neg (P \to Q) $$ + +Recall that implications are also disjunctions: + +$$ P \to Q \equiv \neg P \vee Q $$ + +So we can apply this to our first statement: + +$$ \neg (\neg P \vee Q) $$ + +And De Morgan's law tells us we can "distribute" a negation as long as we change +the disjunction to a conjunction: + +$$ \neg\neg P \wedge \neg Q $$ + +And then we get rid of the double negation: + +$$ P \wedge \neg Q $$ + +And this is our second statement. We have proven that these two statements are +equivalent through this transformation steps as well: + +$$ \neg (P \to Q) \equiv P \wedge \neg Q $$ + +6. + +Q: Simplify the following statements (so that negation only appears right before +variables). + +(a) $\neg (P \to \neg Q)$. + +(b) $(\neg P \vee \neg Q) \to \neg(\neg Q \wedge R)$. + +\(c\) $\neg((P \to \neg Q) \vee (R \wedge \neg R))$. + +(d) It is false that if Sam is not a man then Chris is a woman, and that Chris +is not a woman. + +A: + +(a) $\neg (P \to \neg Q)$. + +$P \to \neg Q$ is false when $P$ is true and $Q$ is false. We can rewrite this +as: + +$$ P \to \neg Q \equiv \neg(P \wedge \neg Q) $$ + +This makes our original statement: + +$$ \neg (P \to \neg Q) \equiv \neg (\neg(P \wedge \neg Q)) $$ + +Let's then use De Morgan's Law: + +$$ \neg (\neg P \vee \neg\neg Q) $$ + +$$ \neg (\neg P \vee Q) $$ + +And again with De Morgan's Law: + +$$ \neg\neg P \wedge \neg Q $$ + +$$ \boxed{P \wedge \neg Q} $$ + +(b) $(\neg P \vee \neg Q) \to \neg(\neg Q \wedge R)$. + +$$ (\neg P \vee \neg Q) \to (\neg\neg Q \vee \neg R) $$ + +$$ (\neg P \vee \neg Q) \to (Q \vee \neg R) $$ + +\(c\) $\neg((P \to \neg Q) \vee (R \wedge \neg R))$. + +$P \to \neg Q$ is false only when $P$ is true and $Q$ is true. + +$$ \neg(\neg(P \wedge Q) \vee (R \wedge \neg R)) $$ + +$$ \neg\neg(P \wedge Q) \wedge \neg(R \wedge \neg R) $$ + +$$ (P \wedge Q) \wedge (\neg R \vee R) $$ + +This looks finished, but we can go one step forward with the second operator: + +$$ \neg R \vee R \equiv T $$ + +This will always be true. + +$$ (P \wedge Q) \wedge T $$ + +Because forming a conjunction with an always true statement will always depend +on the ambiguous statement (the not always true statement), we can simply drop +the always true: + +$$ \boxed{P \wedge Q} $$ + +(d) It is false that if Sam is not a man then Chris is a woman, and that Chris +is not a woman. + +Let $S$ be "Sam is a man" and $C$ be "Chris is a woman". + +$$ \neg (\neg S \to C) \wedge \neg C $$ + +Note the wording here is specific, the "It is false" applies only to the first +part of the problem statement sentence, not the second + +$\neg S \to C$ is false only if $S$ is false and $C$ is false. + +$$ \neg (\neg S \wedge \neg C) \wedge \neg C $$ + +$$ (S \vee C) \wedge \neg C $$ + +We need to consider both "or" cases: + +$$ ((S \wedge C) \vee (S \wedge \neg C)) \wedge \neg C $$ + +$$ (S \wedge (C \vee \neg C)) \wedge \neg C $$ + +$$ (S \wedge T) \wedge \neg C $$ + +$$ \boxed{S \wedge \neg C} $$ diff --git a/chapter_1/1_3/practice_problems.md b/chapter_1/1_3/practice_problems.md new file mode 100644 index 0000000..65753f8 --- /dev/null +++ b/chapter_1/1_3/practice_problems.md @@ -0,0 +1,275 @@ +# 1.3.7 Practice Problems + +1. + +Q: Make a truth table for the statement $(P \wedge Q) \to (P \vee Q)$. + +A: + +| $P$ | $Q$ | $(P \wedge Q)$ | $(P \vee Q)$ | $(P \wedge Q) \to (P \vee Q)$ | +| --- | --- | -------------- | ------------ | ----------------------------- | +| T | T | T | T | T | +| T | F | F | T | T | +| F | T | F | T | T | +| F | F | F | F | T | + +2. + +Q: Complete a truth table for the statement $\neg Q \vee (Q \to P)$. What can +you conclude about $P$ and $Q$ if you knew the statement above was false? + +| $Q$ | $P$ | $\neg Q$ | $Q \to P$ | $\neg Q \vee (Q \to P)$ | +| --- | --- | -------- | --------- | ----------------------- | +| T | T | F | T | T | +| T | F | F | F | F | +| F | T | T | T | T | +| F | F | T | T | T | + +If we know that $\neg Q \to \vee (Q \to P)$ is false, then the only possible +values are $Q = T$ and $P = F$. + +3. Construct a truth table for the statement $Q \to (\neg P \vee R)$. + +| $P$ | $Q$ | $R$ | $\neg P$ | $(\neg P \vee R)$ | $Q \to (\neg P \vee R)$ | +| --- | --- | --- | -------- | ----------------- | ----------------------- | +| T | T | T | F | T | T | +| T | T | F | F | F | F | +| T | F | T | F | T | T | +| T | F | F | F | F | T | +| F | T | T | T | T | T | +| F | F | T | T | T | T | +| F | T | F | T | T | T | +| F | F | F | T | T | T | + +4. + +Q: Determine whether the statements $P \to (Q \vee R)$ and +$(P \to Q) \vee (P \to R)$ are logically equivalent by completing a truth table +for both statements. + +A: + +| $P$ | $Q$ | $R$ | $(Q \vee R)$ | $(P \to Q)$ | $(P \to R)$ | $P \to (Q \vee R)$ | $(P \to Q) \vee (P \to R)$ | +| --- | --- | --- | ------------ | ----------- | ----------- | ------------------ | -------------------------- | +| T | T | T | T | T | T | T | T | +| T | T | F | T | T | F | T | F | +| T | F | T | T | F | T | T | T | +| T | F | F | F | F | F | F | F | +| F | T | T | T | T | T | T | T | +| F | T | F | T | T | T | T | T | +| F | F | T | T | T | T | T | T | +| F | F | F | F | T | T | T | T | + +From this truth table, we can conclude that the statements $P \to (Q \vee R)$ +and $(P \to Q) \vee (P \to R)$ are _not_ logically equivalent statements. + +5. + +Determine if the following is a valid deduction rule: + +$$ +P \to Q \\ +\neg Q \\ +\overline{\therefore \quad \neg P} +$$ + +We can first construct a truth table: + +| $P$ | $Q$ | $\neg Q$ | $P \to Q$ | $\neg P$ | +| --- | --- | -------- | --------- | -------- | +| T | T | F | T | F | +| T | F | T | F | F | +| F | T | F | T | T | +| F | F | T | T | T | + +The last row shows us that this is a valid deduction, where the conclusion, +$\neg Q$ is true and the two premises, $P \to Q$ and $\neg P$ are both true as +well. While this is the only row where the premises are both true, if there were +other rows where the premises were both true, we would then check to see if the +conclusion was true as well. Only if all the rows all had the premises as being +true as well as the conclusion being true could we say that this is a valid +deduction. + +Since we only have one row where both premises are true though, we only check +this one, which is valid. This is a valid deduction. + +6. Determine if the following is a valid deduction rule: + +$$ +P \to (Q \vee R) \\ +\neg (P \to Q) \\ +\overline{\therefore \quad R} +$$ + +| $P$ | $Q$ | $R$ | $(Q \vee R)$ | $(P \to Q)$ | $P \to (Q \vee R)$ | \neg(P \to Q) | +| --- | --- | --- | ------------ | ----------- | ------------------ | ------------- | +| T | T | T | T | T | T | F | +| T | T | F | T | T | T | F | +| T | F | T | T | F | T | T | +| T | F | F | F | F | F | T | +| F | T | T | T | T | T | F | +| F | T | F | T | T | T | F | +| F | F | T | T | T | T | F | +| F | F | F | F | T | T | F | + +Let's now isolate the rows where the premises $P \to (Q \vee R)$ and +$\neg (P \to Q)$ are both true: + +| $P$ | $Q$ | $R$ | $(Q \vee R)$ | $(P \to Q)$ | $P \to (Q \vee R)$ | \neg(P \to Q) | +| --- | --- | --- | ------------ | ----------- | ------------------ | ------------- | +| T | F | T | T | F | **T** | **T** | + +Now let's see if within these remaining rows if $R$ is true: + +| $P$ | $Q$ | $R$ | $(Q \vee R)$ | $(P \to Q)$ | $P \to (Q \vee R)$ | \neg(P \to Q) | +| --- | --- | ----- | ------------ | ----------- | ------------------ | ------------- | +| T | F | **T** | T | F | **T** | **T** | + +Since the only row where both premises $P \to (Q \vee R)$ and $\neg(P \to Q)$ +are true also has $R$ true, this is a valid deduction rule. + +7. Determine if the following is a valid deduction rule: + +$$ +(P \wedge Q) \to R \\ +\neg P \vee \neg Q \\ +\overline{\therefore \neg R} +$$ + +| $P$ | $Q$ | $R$ | $(P \wedge Q)$ | $(P \wedge Q) \to R$ | $\neg P$ | $\neg Q$ | $\neg P \vee \neg Q$ | $\neg R$ | +| --- | --- | --- | -------------- | -------------------- | -------- | -------- | -------------------- | -------- | +| T | T | T | T | T | F | F | F | F | +| T | T | F | T | F | F | F | F | T | +| T | F | T | F | T | F | T | T | F | +| T | F | F | F | T | F | T | T | T | +| F | T | T | F | T | T | F | T | F | +| F | T | F | F | T | T | F | T | T | +| F | F | T | F | T | T | T | T | F | +| F | F | F | F | T | T | T | T | T | + +Let's now isolate the rows where the premises $(P \wedge Q) \to R$ and +$\neg P \vee \neg Q$ are both true: + +| $P$ | $Q$ | $R$ | $(P \wedge Q)$ | $(P \wedge Q) \to R$ | $\neg P$ | $\neg Q$ | $\neg P \vee \neg Q$ | $\neg R$ | +| --- | --- | --- | -------------- | -------------------- | -------- | -------- | -------------------- | -------- | +| T | F | T | F | T | F | T | T | F | +| T | F | F | F | T | F | T | T | T | +| F | T | T | F | T | T | F | T | F | +| F | T | F | F | T | T | F | T | T | +| F | F | T | F | T | T | T | T | F | +| F | F | F | F | T | T | T | T | T | + +Now let's see if within these remaining rows if all values for $\neg R$ are +true...they are not, so this is not a valid deduction rule. + +8. Determine if the following is a valid deduction rule: + +$$ +P \to Q \\ +P \wedge \neg Q \\ +\overline{\therefore R} +$$ + +| $P$ | $Q$ | $R$ | $\neg Q$ | $P \to Q$ | $P \wedge \neg Q$ | +| --- | --- | --- | -------- | --------- | ----------------- | +| T | T | T | F | T | F | +| T | T | F | F | T | F | +| T | F | T | T | F | T | +| T | F | F | T | F | T | +| F | T | T | F | T | F | +| F | T | F | F | T | F | +| F | F | T | T | T | F | +| F | F | F | T | T | F | + +Now, let's isolate the rows where the premises $P \to Q$ and $P \wedge \neg Q$ +are true: + +| $P$ | $Q$ | $R$ | $\neg Q$ | $P \to Q$ | $P \wedge \neg Q$ | +| --- | --- | --- | -------- | --------- | ----------------- | +| T | T | T | F | T | F | +| T | T | F | F | T | F | +| F | T | T | F | T | F | +| F | T | F | F | T | F | +| F | F | T | T | T | F | +| F | F | F | T | T | F | + +As you can see, in the remaining rows after isolating all rows where $P \to Q$ +is true, all rows for $P \wedge \neg Q$ are false. In order to have a valid +deduction rule, all premises must first be true, and then the corresponding +conclusion must also be true. Since both premises are not valid in every +remaining case, there is no case to test the conclusion against. No +counterexamples exist. Since there are no cases where all premises are true but +the conclusion is false, this means that the given statement is +[vacuously valid](https://en.wikipedia.org/wiki/Vacuous_truth) and is therefore +a valid deduction rule. + +9. + +Q: Which of the following statements is a _law of logic_? That is, which of the +following are true no matter what your domain of discourse is and no matter what +you interpret the predicates as meaning? Select all that apply. + +A: $\forall x (P(x) \vee \neg P(x))$. + +B. $\exists x P(x) \to \forall x P(x)$. + +C. $\neg \forall x P(x) \to \exists x P(x)$. + +D. $\forall x \exists y P(x, y) \leftrightarrow \exists y \forall x P(x, y)$. + +A: + +A law of logic is a statement in predicate logic that is necessarily true. + +A: $\forall x (P(x) \vee \neg P(x))$. + +This statement does follow the law of logic. It is saying "For all things $x$, +some fact $P(x)$ is true or that same fact, $P(x)$ is not true." In extremely +basic English, "All things are either something or not something." + +B. $\exists x P(x) \to \forall x P(x)$. + +This statement does _not_ follow the law of logic. It is saying "If there exists +some thing, $x$, for which there is some truth $P(x)$, then for all things that +are $x$, that truth $P(x)$ is true." This is not necessarily true. + +C. $\neg \forall x P(x) \to \exists x P(x)$. + +This statement does _not_ follow the law of logic. It is saying "For all things +$x$, if $P(x)$ is not true, then there exists at least one $x$ for which $P(x)$ +is true." A more accurate statement would be: + +$$ \neg \forall x P(x) \to \exits x \neg P(x) $$ + +This one tripped me up, so for notations sake, let's say $P(x)$ means "person +$x$ cares". So the statement given in C would become "If nobody cares, then +there exists at least someone who cares." Which isn't true, there could truly be +0 people who care. + +The last statement I gave that is more accurate would be saying "If nobody +cares, then there exists at least one person who doesn't care." + +D. $\forall x \exists y P(x, y) \leftrightarrow \exists y \forall x P(x, y)$. + +"For all things $x$, there is at least one $y$ where $P(x, y)$ is true if and +only if there exists at least one $y$ for every $x$ that $P(x, y)$ is true." + +These do not follow the law of logic, but to understand it's best to define $x$ +and $y$ in some context. + +Let $x$ be "boss" and $y$ be "employee", and $P(x, y)$ be "boss is an employer +of an employee." + +The first statement would be: + +"For all bosses, there exists at least one employee for which the boss is an +employer of an employee." + +The second statement would be: + +"There exists at least one employee for all bosses for which the boss is an +employer of an employee." + +The first statement is true, but the second statement is false, an "if and only +if" statement is true only if both statements are true or both statements are +false. diff --git a/chapter_1/1_3/preview_activity.md b/chapter_1/1_3/preview_activity.md index a0b0f3d..fe7d248 100644 --- a/chapter_1/1_3/preview_activity.md +++ b/chapter_1/1_3/preview_activity.md @@ -22,7 +22,7 @@ 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 $\to \neg B$. +($P \wedge G$), then he chooses to not wear a bow tie $\to \neg B$. B. $P \wedge G \to B$ @@ -139,7 +139,7 @@ $$ P \to Q = T \to F = F $$ $$ \neg P \vee Q = F \vee F = F $$ -Both are false, so this is not a case where both statements are true.. +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. diff --git a/chapter_1/1_3/reading_questions.md b/chapter_1/1_3/reading_questions.md new file mode 100644 index 0000000..86d9afe --- /dev/null +++ b/chapter_1/1_3/reading_questions.md @@ -0,0 +1,36 @@ +# 1.3.6 Reading Questions + +1. + +Q: To check whether two statements are logically equivalent, you can use a truth +table. Explain what you would look for in the truth table to conclude that the +two statements are logically equivalent. What would tell you they are _not_ +logically equivalent? + +A: + +If the two statements always result in the same corresponding conclusions. In +other words, one one statement's conclusion is true, then so is the second +statement's conclusion, and similarly if one statement's conclusion is false, so +is the second statement's conclusion. This would be the case where they are +logically equivalent. + +If one of the two statements' resulting conclusions do not correspond with each +other, then the two statements are not logically equivalent. For example, if +even one example could be given where one statement's conclusion was true, but +the other statement's conclusion was false, this would invalidate the two +statements' logical equivalency. + +This is done by checking their rows in a truth table. + +2. + +Q: To check whether a deduction rule is _valid_, you can use a truth table. +Explain what you would look for in the completed truth table to say that the +deduction rule is valid, and what would tell you the deduction rule is _not_ +valid. + +A: To check whether a deduction rule is valid using a truth table, we look at +all rows where every premise is true. If in every such row the conclusion is +also true, then the rule is valid. If there is at least one row where all the +premises are true but the conclusion is false, then the rule is not valid. diff --git a/leftoff.txt b/leftoff.txt index 987e7ca..85322d0 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -77 +79