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@ -496,3 +496,479 @@ $$ (S \wedge (C \vee \neg C)) \wedge \neg C $$
$$ (S \wedge T) \wedge \neg C $$ $$ (S \wedge T) \wedge \neg C $$
$$ \boxed{S \wedge \neg C} $$ $$ \boxed{S \wedge \neg C} $$
7.
Q: Use De Morgan's Laws and any other logical equivalence facts you know to
simplify the following statements. Show all your steps. Your final statements
should have negations only appear directly next to the sentence variables or
predicates ($P$, $Q$, $E(x)$, etc.), and no double negations. It would be a good
idea to use only conjunctions, disjuctions, and negations.
(a) $\neg((\neg P \wedge Q) \vee \neg (R \vee \neg S))$.
$$ \neg((\neg P \wedge Q) \vee \neg (R \vee \neg S)) $$
Let's "distribute" the outside negation first:
$$ \neg(\neg P \wedge Q) \wedge \neg\neg (R \vee \neg S) $$
Get rid of the double negative on the second "term":
$$ \neg(\neg P \wedge Q) \wedge (R \vee \neg S) $$
And "distribute" the negation on the first "term":
$$ (\neg\neg P \vee \neg Q) \wedge (R \vee \neg S) $$
And again, get rid of the double negative:
$$ \boxed{(P \vee \neg Q) \wedge (R \vee \neg S)} $$
(b) $\neg ((\neg P \to \neg Q) \wedge (\neg Q \to R))$ (careful with the
implications)
$$ \neg ((\neg P \to \neg Q) \wedge (\neg Q \to R)) $$
First let's "distribute" the outside negation:
$$ \neg(\neg P \to \neg Q) \vee \neg(\neg Q \to R) $$
Now apply "Implications are Disjunctions":
$$ \neg(P \vee \neg Q) \vee \neg(Q \vee R) $$
Note the reason it takes this form is that implications are disjunctions only
negates the antecedent (the first term of an if/then statement, not the
consequent). Now we can apply De Morgan's Laws again to each "term":
$$ (\neg P \wedge Q) \vee (\neg Q \wedge \neg R) $$
\(c\) For both parts above, verify your answers are correct using truth tables.
That is, use a truth table to check that the given statement and your proposed
simplification are actually logically equivalent.
For part (a), we start at: $\neg((\neg P \wedge Q) \vee \neg (R \vee \neg S))$,
and we end at $(P \vee \neg Q) \wedge (R \vee \neg S)$:
| $P$ | $Q$ | $R$ | $S$ | $\neg((\neg P \wedge Q) \vee \neg(R \vee \neg S))$ | $(P \vee \neg Q) \wedge (R \vee \neg S)$ |
| --- | --- | --- | --- | -------------------------------------------------- | ---------------------------------------- |
| T | T | T | T | T | T |
| T | T | T | F | T | T |
| T | T | F | T | F | F |
| T | T | F | F | T | T |
| T | F | T | T | T | T |
| T | F | T | F | T | T |
| T | F | F | T | F | F |
| T | F | F | F | F | F |
| F | T | T | T | F | F |
| F | T | T | F | F | F |
| F | T | F | T | F | F |
| F | T | F | F | F | F |
| F | F | T | T | T | T |
| F | F | T | F | T | T |
| F | F | F | T | F | F |
| F | F | F | F | T | T |
These two statements are logically equivalent, as their truth tables, when
evaluated, produce exactly the same results in all cases. It is reasonable to
conclude that:
$$ \neg((\neg P \wedge Q) \vee \neg(R \vee \neg S)) \equiv (P \vee \neg Q) \wedge (R \vee \neg S) $$
8.
Q: Consider the statement, "If a number is triangular or square, then it is not
prime"
(a) Make a truth table for the statement $(T \vee S) \to \neg P$.
(b) If you believed the statement was _false_, what properties would a
counterexample need to possess? Explain by referencing your truth table.
\(c\) If the statement were true, what could you conclude about the number 5657,
which is definitely prime? Again, explain using the truth table.
A:
(a) Make a truth table for the statement $(T \vee S) \to \neg P$.
| $T$ | $S$ | $P$ | $(T \vee S)$ | $\neg P$ | $(T \vee S) \to \neg P$ |
| --- | --- | --- | ------------ | -------- | ----------------------- |
| T | T | T | T | F | F |
| T | T | F | T | T | T |
| T | F | T | T | F | F |
| T | F | F | T | T | T |
| F | T | T | T | F | F |
| F | T | F | T | T | T |
| F | F | T | F | F | T |
| F | F | F | F | T | T |
(b) If you believed the statement was _false_, what properties would a
counterexample need to possess? Explain by referencing your truth table.
The only examples where the statement $(T \vee S) \to \neg P$ are false are
indicated by these rows from the truth table:
| $T$ | $S$ | $P$ | $(T \vee S)$ | $\neg P$ | $(T \vee S) \to \neg P$ |
| --- | --- | --- | ------------ | -------- | ----------------------- |
| T | T | T | T | F | F |
| T | F | T | T | F | F |
| F | T | T | T | F | F |
From the truth table, $(T \vee S) \to \neg P$ is false exactly when:
- $T \vee S = T$ (the number is triangular or square, or both), and
- $\neg P = F$, meaning $P = T$ (the number is prime).
So a counterexample must be a prime number that is either triangular, square, or
both.
In other words, the number must satisfy:
$$ P = T \quad \text{ and } \quad (T \vee S) = T $$
Equivalently, the number is prime but also has at least one of the properties
"triangular" or "square."
\(c\) If the statement were true, what could you conclude about the number 5657,
which is definitely prime? Again, explain using the truth table.
Let us express this as $P(5657) = T$, and also the statement itself is true,
$(T \vee S) \to \neg P = T$.
If we isolate our table where $P = T$ and $(T \vee S) \to \neg P = T$, then we
get:
| $T$ | $S$ | $P$ | $(T \vee S)$ | $\neg P$ | $(T \vee S) \to \neg P$ |
| --- | --- | --- | ------------ | -------- | ----------------------- |
| F | F | T | F | F | T |
This implies that since 5657 is prime, and the statement is true, that the
hypothesis $(T \vee S)$ in this case is false, while the conclusion, $\neg P$ is
true. Therefore, 5657 is neither triangular nor square.
9.
Q: Tommy Flanagan was telling you what he ate yesterday afternoon. He tells you,
"I had either popcorn or raisins. Also, if I had cucumber sandwiches, then I had
soda. But I didn't drink soda or tea." Of course, you know that Tommy is the
world's worst liar, and everything he says is false. What did Tommy eat?
Justify your answer by writing all of Tommy's statements using sentence
variables ($P$, $Q$, $R$, $S$, $T$), taking their negations, and using these to
deduce what Tommy actually ate.
A:
Let the following variables represent the following statements:
$P$: "Tommy had popcorn."
$Q$: "Tommy had raisins."
$R$: "Tommy had cucumber sandwiches."
$S$: "Tommy had soda."
$T$: "Tommy had tea."
Breaking this down for later use, we get:
$(P \vee Q)$: "Tommy had either popcorn or raisins."
$R \to S$ "If Tommy had cucumber sandwiches, then he had soda."
$\neg(S \vee T)$: "Tommy doesn't drink soda or tea."
This evaluates to:
$$ (P \vee Q) \wedge (R \to S) \wedge \neg(S \vee T) $$
And lastly we know Tommy is a liar, so we negate the entire statement:
$$ \neg((P \vee Q) \wedge (R \to S) \wedge \neg(S \vee T)) $$
Let's use De Morgan's Laws and Implications are Disjunctions and negations to
see if we can't express this in a way that we can find out what Tommy ate.
First, implications are disjunctions for the "middle term":
$$ \neg((P \vee Q) \wedge (\neg R \vee S) \wedge \neg(S \vee T)) $$
Let's apply De Morgan's Laws to the "last term":
$$ \neg((P \vee Q) \wedge (\neg R \vee S) \wedge (\neg S \wedge \neg T)) $$
And De Morgan's Laws again on the entire statement:
$$ \neg(P \vee Q) \vee \neg(\neg R \vee S) \vee \neg(\neg S \wedge \neg T) $$
$$ (\neg P \wedge \neg Q) \vee (R \wedge \neg S) \vee (S \vee T) $$
Ultimately now we have to prove that this statement will always be false, since
Tommy is a liar.
$$ (\neg P \wedge \neg Q) \vee (R \wedge \neg S) \vee (S \vee T) = F $$
Since these are three separate atomic statements joined by disjunctions, this
means that each individual statement must be false.
$$ (\neg P \wedge \neg Q) = F $$
So this means that Tommy did not have popcorn or raisins, but he might have had
one or the other.
$$ (R \wedge \neg S) = F $$
This means that Tommy did not have cucumber sandwiches, but he did have a soda,
or he did have cucumber sandwiches and did not have a soda.
$$ (S \vee T) = F $$
This means that Tommy didn't drink Soda or Tea. He didn't drink anything. This
is the most definitive statement. Knowing this, we can further refine our
evaluation of the other two statements.
$$ (R \wedge \neg S) = F $$
Since we know that $\neg S = T$, this means that $R = F$. Tommy did not eat
cucumber sandwiches.
$$ (\neg P \wedge \neg Q) = F $$
Here we know the least, we know that Tommy either had popcorn or raisins, but he
had at least one of them.
In conclusion, we know the following:
- Tommy did not drink soda or tea
- Tommy did not eat cucumber sandwiches
- Tommy either had either popcorn or raisins or both.
10. Can you chain implications together? That is, if $P \to Q$ and $Q \to R$,
does that mean the $P \to R$? Prove that the following is a valid deduction
rule:
$$
P \to Q \\
Q \to R \\
\overline{\therefore P \to R}
$$
This is solved best when using a truth table:
| $P$ | $Q$ | $R$ | $P \to Q$ | $Q \to R$ | $P \to R$ |
| --- | --- | --- | --------- | --------- | --------- |
| T | T | T | T | T | T |
| T | T | F | T | F | F |
| T | F | T | F | T | T |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | T | F | T | F | T |
| F | F | T | T | T | T |
| F | F | F | T | T | T |
We're only concerned where our antecedents are true, so let's highlight which
ones have antecedents that are both true, wherever this is the case, we will
also highlight the conclusion's true/false value and compare it to the
antecedents.
| $P$ | $Q$ | $R$ | $P \to Q$ | $Q \to R$ | $P \to R$ |
| --- | --- | --- | --------- | --------- | --------- |
| T | T | T | **T** | **T** | **T** |
| T | T | F | T | F | F |
| T | F | T | F | T | T |
| T | F | F | F | T | F |
| F | T | T | **T** | **T** | **T** |
| F | T | F | T | F | T |
| F | F | T | **T** | **T** | **T** |
| F | F | F | **T** | **T** | **T** |
And as we can see, when we do this, the consequent is also always true,
therefore this is a valid deduction rule.
11.
Q: Suppose $P$ and $Q$ are (possibly molecular) propositional statements. Prove
that $P$ and $Q$ are logically equivalent if and only if $P \leftrightarrow Q$
is a tautology.
A:
Let's break this down with the standard "if and only if" truth table:
| $P$ | $Q$ | $P \leftrightarrow Q$ |
| --- | --- | --------------------- |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
For $P$ and $Q$ to be logically equivalent, both of the following statements
must be true:
$$ P \to Q $$
And
$$ Q \to P $$
This evaluates to:
$$ (P \to Q) \wedge (Q \to P) $$
Let's expand our truth table to include these statements as well so we can
further evaluate whether or not $P \leftrightarrow Q$ is a tautology:
| $P$ | $Q$ | $P \to Q$ | $Q \to P$ | $(P \to Q) \wedge (Q \to P)$ | $P \leftrightarrow Q$ |
| --- | --- | --------- | --------- | ---------------------------- | --------------------- |
| T | T | T | T | T | T |
| T | F | F | T | F | F |
| F | T | T | F | F | F |
| F | F | T | T | T | T |
As seen in the last two columns, $(P \to Q) \wedge (Q \to P)$ and
$P \leftrightarrow Q$ have identical truth values in every row. This means they
are logically equivalent. Since $P \leftrightarrow Q$ is true exactly when $P$
and $Q$ have the same truth value, it follows that it is true in all cases where
logical equivalence holds, and thus characterizes a tautology condition.
12.
Q: Suppose $P_1, P_2, \dots , P_n$ and $Q$ are (possibly molecular)
propositional statements. Suppose further that
$$
P_1 \\
P_2 \\
\vdots \\
P_n \\
\overline{\therefore Q}
$$
is a valid deduction rule. Prove that the statement
$$ (P_1 \wedge P_2 \wedge \dots \wedge P_n) \to Q $$
is a tautology.
A:
A deduction rule is valid where there are no premises that result in a false
conclusion. Since we know that:
$$
P_1 \\
P_2 \\
\vdots \\
P_n \\
\overline{\therefore Q}
$$
is a valid deduction rule, we also know then that there is no case where all
$P_1 \dots P_n$ are true and $Q$ is false. The implication is that
$(P_1 \wedge \dots \wedge P_n) \to Q$ is always true, therefore it is a
tautology.
$$ (P_1 \wedge \dots \wedge P_n) \to Q $$
13.
Q: Consider the statements below. Translate each into symbols, using the
predicate $F(x, y)$ for "person $x$ can be fooled at any time $y$." Decide
whether any of the statements are equivalent to each other, or whether any imply
any others, in this context or in general.
(a) You can fool some people all of the time.
(b) You can fool everyone some of the time.
\(c\) You can always fool some people.
(d) Sometimes you can fool everyone.
A:
Let's first just translate each of these into symbols
(a) You can fool some people all of the time.
$$ \exists x \forall y F(x, y) $$
(b) You can fool everyone some of the time.
$$ \forall x \exists y F(x, y) $$
\(c\) You can always fool some people.
$$ \forall y \exists x F(x, y) $$
(d) Sometimes you can fool everyone.
$$ \exists y \forall x F(x, y) $$
None of these statements are logically equivalent to each other. Equivalencies
with quantifiers always involve some negation of the quantifier(s) or the
antecedent(s). There are no negations here and so none of them are logically
equivalent. However yes, some do imply others.
(a) does imply \(c\):
$$ \exists x \forall y F(x, y) \to \forall y \exists x F(x, y) $$
"If there exists some person $x$ for all times $y$ for which that person is
fooled, then for all times $y$, there is at least one person $x$ who is fooled."
and (d) does imply (b):
$$ \exists y \forall x F(x, y) \to forall x \exists y F(x, y) $$
"If there exists some time $y$ for all people $x$ where all people are fooled,
then for all people $x$ there exists at least one time $y$ where all people are
fooled."
14.
Q: Suppose $P(x)$ is some predicate for which the statement $\forall x P(x)$ is
true. Is it also the case that $\exists x P(x)$ is true? In other words, is the
statement $\forall x P(x) \to \exists x P(x)$ always true? Is the converse
always true? Assume the domain of discourse is non-empty.
A:
Let's break this down:
Suppose $P(x)$ is some predicate for which the statement $\forall x P(x)$ is
true. Is it also the case that $\exists x P(x)$ is true?
Is this true?:
$$ \forall x P(x) \to \exists x P(x) $$
Since we know that $\forall x P(x)$ is true for all $x$, $P(x)$ holds for every
element of the domain. Because we also know that the domain of discourse is not
empty, we can choose some element $a$ for which $P(a)$ is true. Therefore,
$\exists x P(x)$ is true.
15.
Q: Simplifying negations will be especially useful when we try to prove a
statement by considering what would happen if it were false. For each statement
below, write the _negation_ of the statement as simply as possible. Don't just
say, "It is false that..."
(a) Every number is either even or odd.
(b) There is a sequence that is both arithmetic and geometric.
\(c\) For all numbers, $n$, if $n$ is prime, then $n + 3$ is not prime.
A:
(a) Every number is either even or odd.
(b) There is a sequence that is both arithmetic and geometric.
\(c\) For all numbers, $n$, if $n$ is prime, then $n + 3$ is not prime.

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# Example 1.3.13
Can you switch the order of quantifiers? For example, consider the two
statements:
$$ \forall x \exists y P(x, y) \quad \text{ and } \quad \exists y \forall x P(x, y) $$
Are these logically equivalent?
**Solution**.
These statements are NOT logically equivalent. To see this, we should provide an
interpretation of the predicate $P(x, y)$ which makes one of the statements true
and the other false.
Let $P(x, y)$ be the predicate $x < y$. It is true, in the natural numbers, that
for all $x$ there is some $y$ greater than that $x$ (since there are infinitely
many numbers). However, there is no natural number $y$ which is greater than
every number $x$. Thus it is possible for $\forall x \exists y P(x, y)$ to be
true while $\exists y \forall x P(x, y)$ is false.
We cannot do the reverse of this though. If there is some $y$ for which every
$x$ satisfies $P(x, y)$, then certainly for every $x$ there is some $y$ which
satisfies $P(x, y)$. The first is saying we can find one $y$ that works for
every $x$. The second allows different $y$'s to work for different $x$'s, but
nothing is preventing us from using the same $y$ that works for every $x$. In
other words, while we don't have logical equivalence between the two statements,
we do have a valid deduction rule:
$$
\exists y \forall x P(x, y) \\
\overline{\therefore \forall x \exists y P(x, y)}
$$
Put yet another way, this says that the single statement
$$ \exists y \forall x P(x, y) \to \forall x \exists y P(x, y) $$
is always true; it is a law of logic.