The fully graded problems are 1.4 #50 and 1.6 #35. Complete solutions to the homework are below. These solutions are not unique; many of the problems can be satisfactorily answered in more than one way.
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1.1 #14
a) $r\wedge\neg q$
b) $p\wedge q\wedge r$
c) $r\to p$
d) $p\wedge\neg q\wedge r$
e) $(p\wedge q)\to r$
f) $r\leftrightarrow(p\vee q)$
1.1 #16
a) true, b) false, c) true, d) false
1.1 #28
Only part (a) is done here, as an example:
Converse: If I ski tomorrow, then it snowed today.
Contrapositive: If I don’t ski tomorrow, then it didn’t snow today.
Inverse: If it doesn’t snow today, then I won’t ski tomorrow.
1.1 #31 – Check back of book
1.3 #22
$$\begin{array}{c c c | c | c | c | c | c} p & q & r & p\to q & p\to r & (p\to q)\wedge(p\to r) & q\wedge r & p\to(q\wedge r) \\ \hline T&T&T&T&T&T&T&T \\ T&T&F&T&F&F&F&F \\ T&F&T&F&T&F&F&F \\ T&F&F&F&F&F&F&F \\ F&T&T&T&T&T&T&T \\ F&T&F&T&T&T&F&T \\ F&F&T&T&T&T&F&T \\ F&F&F&T&T&T&F&T \end{array}$$
Because the columns for $(p\to q)\wedge(p\to r)$ and $p\to(q\wedge r)$ are identical, those propositions are logically equivalent.
Note that $(p\to q)\wedge(p\to r)$ asserts that $p$ implies $q$ and that $p$ implies $r$, while $p\to(q\wedge r)$ asserts that $p$ implies $q$ and $r$. It should seem very plausible that these assertions are equivalent; the truth table verifies that this is the case.
1.3 #30
$$\begin{array}{c c c * 6 {| c}} p & q & r & p\vee q & \neg p & \neg p\vee r & (p\vee q)\wedge(\neg p\vee r) & q\vee r & (p\vee q)\wedge(\neg p\vee r) \to q\vee r \\ \hline T&T&T&T&F&T&T&T&T \\ T&T&F&T&F&F&F&T&T \\ T&F&T&T&F&T&T&T&T \\ T&F&F&T&F&F&F&F&T \\ F&T&T&T&T&F&F&T&T \\ F&T&F&T&T&F&F&T&T \\ F&F&T&F&T&F&F&T&T \\ F&F&F&F&T&F&F&F&T \end{array}$$
The last column is all T’s, so $(p\vee q)\wedge(\neg p\vee r) \to q\vee r$ is a tautology.
We can also reason this problem out “in plain English”. To show that $(p\vee q)\wedge(\neg p\vee r)\to(q\vee r)$ is a tautology, it suffices to show that when $(p\vee q)\wedge(\neg p\vee r)$ is true, $q\vee r$ must also be true. Suppose $(p\vee q)\wedge(\neg p\vee r)$ is true. Then both $p\vee q$ and $\neg p\vee r$ are true. That is, $p$ or $q$ is true, and $\neg p$ or $r$ is true. But $p$ and $\neg p$ can’t both be true, so either $p$ is false (forcing $q$ to be true) or $\neg p$ is false (forcing $r$ to be true). Therefore, at least one of $q,r$ is true, and so $q\vee r$ is true. This completes the proof that $(p\vee q)\wedge(\neg p\vee r)\to(q\vee r)$ is a tautology.
1.3 #32
To show that two compound propositions are not equivalent, it suffices to find an assignment of truth values to the propositional variables that makes one of the propositions true and the other false.
If $p$ is true, $q$ is false, and $r$ is false, then $p\wedge q$ is false, so $(p\wedge q)\to r$ is vacuously true. However, $p\to r$ is false, so $(p\to r)\wedge(q\to r)$ is false. Thus $(p\wedge q)\to r$ is not logically equivalent to $(p\to r)\wedge(q\to r)$.
1.3 #40
$p\wedge q\wedge\neg r$
1.4 #8
a) Every rabbit hops.
b) Every animal is a rabbit that hops.
c) The following translations are all equivalent (the first is the most direct reading, but the least clear as an English sentence):
- There exists an animal such that if it is a rabbit, it hops.
- There exists an animal that either hops or is not a rabbit.
- Not all animals are rabbits that don’t hop.
d) There exists a rabbit that hops.
$\bigstar$ 1.4 #50
For example, let $x$ be drawn from the domain of integers, let $P(x)$ be the statement “$x$ is even”, and let $Q(x)$ be the statement “$x$ is odd”. Then $\forall x P(x)\vee\forall x Q(x)$ asserts that either all integers are even or all integers are odd. That is false. But $\forall x (P(x)\vee Q(x))$ asserts that every integer is either even or odd, which is true. So, these assertions are not equivalent.
1.4 #60
a) $\forall x (P(x)\to Q(x))$
b) $\exists x (R(x)\wedge\neg Q(x))$
c) $\exists x (R(x)\wedge\neg P(x))$
d) Yes. From premise (b), some excuses are unsatisfactory. From premise (a), using the rule universal modus tollens (p. 77 of the textbook), we infer that those excuses which are unsatisfactory are not clear explanations.
1.5 #8
a) $\exists x~\exists y ~Q(x,y)$
b) $\neg\exists x~\exists y ~Q(x,y)$
c) $\exists x~[Q(x,\textit{Jeopardy!}) \wedge Q(x,\textit{Wheel of Fortune})]$
d) $\forall y~\exists x ~Q(x,y)$
e) $\exists x~\exists y ~[Q(x,\textit{Jeopardy!}) \wedge Q(y,\textit{Jeopardy!}) \wedge (x\ne y)]$
1.5 #28
a) True. (For every $x$, it is possible to calculate $x^2$ and call the result $y$.)
b) False. (Counterexample: if $x=-1$, then there is no real $y$ such that $x=y^2$.)
c) True. (Let $x=0$. Then, for every $y$, we have $xy=0$.)
d) False. (There are no $x,y$ for which $x+y\ne y+x$.)
e) True. (For every real $x$, if $x\ne 0$, then $x$ has a multiplicative inverse $1/x$.)
f) False. (The statement says that there is some $x$ such that all $y$ except 0 are multiplicative inverses of $x$. That is not true.)
g) True. (Given any $x$, let $y=1-x$; then $x+y=1$.)
h) False. (The system of equations has no solution, since $x+2y=2$ implies that $2x+4y=4\ne 5$.)
i) False. (Counterexample: If $x=0$, then the predicate becomes $y=2\wedge -y=1$. There is no $y$ satisfying these conditions.)
j) True. (For all $x,y$, it is possible to evaluate $(x+y)/2$ and call the result $z$.)
1.6 #10
a) Conclusion: I did not play hockey the previous day.
Reasoning: Let $h$ denote “I play hockey [at day $n$]“. Let $s$ denote “I am sore [at day $n+1$]“. Let $w$ denote “I use the whirlpool [at day $n+1$]“. The premises of the argument are $h\to s$, $s\to w$, and $\neg w$. By hypothetical syllogism (p. 72), we infer that $h\to s$ and $s\to w$ imply $h\to w$. By modus tollens (i.e. modus ponens + equivalence of contrapositives), we infer that $h\to w$ and $\neg w$ imply $\neg h$.
b) Possible conclusions include: It was partly sunny on Tuesday or I did not work on Tuesday. It was sunny on Friday or I did not work on Friday. It was sunny on Friday or I worked on Monday.
c) Possible conclusions include: Spiders are not insects. Some non-insects eat some insects. [and various others]
d) Conclusion: Homer is not a student.
Reasoning: The first premise can be phrased as “If person $x$ is a student, then person $x$ has an Internet account.” Homer does not have an account; thus, by universal modus tollens, Homer is not a student. Note that we cannot determine whether Maggie is a student or not.
e) Possible conclusions include: All foods that taste good are not healthy to eat. Tofu does not taste good. All foods that you eat are not healthy to eat.
“You do not eat tofu” is given as a premise, but even if it weren’t, it could be inferred from the other premises. It does not appear that any interesting conclusions can be drawn about cheeseburgers (in particular, it cannot be inferred that you eat them, nor that they taste good).
f) Conclusion: I see elephants running down the road.
Reasoning: Let $d=$ “I am dreaming”, $h=$ “I am hallucinating”, $e=$ “I see elephants running down the road”. The premises are $d\vee h$, $\neg d$, and $h\to e$. By the rule of disjunctive syllogism, $d\vee h$ and $\neg d$ imply $h$. By modus ponens, $h$ and $h\to e$ imply $e$.
$\bigstar$ 1.6 #35
Let $A=$ “Superman is able to prevent evil”, $W=$ “Superman is willing to prevent evil”, $P=$ “Superman prevents evil”, $I=$ “Superman is impotent”, $M=$ “Superman is malevolent”, and $E=$ “Superman exists”.
The premises are:
- $(A\wedge W)\to P$
- $\neg A\to I$
- $\neg W\to M$
- $\neg P$
- $E\to (\neg I\wedge\neg M)$
The conclusion is $\neg E$. This is valid; here’s why. From (1) and (4), we infer $\neg(A\wedge W)$ (modus tollens). By de Morgan’s laws, this is equivalent to $\neg A\vee\neg W$. Either $\neg A$ or $\neg W$ holds; thus, from (2) and (3), either $I$ or $M$ follows (modus ponens). So we can infer $I\vee M$. By de Morgan’s laws, this is equivalent to $\neg(\neg I\wedge\neg M)$. Finally, from (5), we infer $\neg E$ (modus tollens again).