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# 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:
$$ ((S \wedge C) \vee (S \wedge \neg C)) \wedge \neg C $$
$$ (S \wedge (C \vee \neg C)) \wedge \neg C $$
$$ (S \wedge T) \wedge \neg C $$
$$ \boxed{S \wedge \neg C} $$

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# 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.

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@ -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.

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@ -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.

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@ -1 +1 @@
77
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