Matti's CS Notebook

• Discrete mathematics and its applications, Rosen (2019)
• 1. The foundations logic and proofs
• 1.1 Propositional logic

Problems 26-54

26

(a) “If I will remember to send you the address, then you send me an e-mail message.”

(b) “If you were born in the United States, then one is a citizen.”

(c) “If you keeo your textbook, then ti will be a useful reference in your future courses.”

(d) “If the goalie plays well, then the Red Wings will win Stanley Cup.”

(e) “If you get the job, then you had the credentials.”

(f) “If there is a storm, then the beach erodes.”

(g) “If one has a valid password, then it’s possible to log on to the server.”

(h) “If you begin your climb early, you will reach the summit”.

(i) “If you are among the first $100$ customers tomorrow, then you will get a free ice cream cone”.

27

(a) “You buy an ice cream cone, if and only if it is hot ouside.”

(b) “You win the contest if and only if you have the only winning ticket.”

(c) “You get promoted if and only if you have connections”.

(d) “If you watch television if and only if your mind will decay.”

(e) “The trains run late if and only if I take it.”

28

(a) “You get an A in this course if and only if you learn how to solve discrete mathematics problems.”

(b) “If you read the newspaper every day if and only if you will be informed.”

(c) “It rains if and only if it is a weekend day”.

(d) “You can see the wizard if and only if the wizard is not in.”

(e) “My airplane flight is late if and only if I have to catch a connecting flight”.

29 Given a conditional statement of the form $p \rightarrow q$, the converse of the proposition is $q \rightarrow p$, the contrapositive is $\lnot q \rightarrow \lnot p$, and the inverse is $\lnot p \rightarrow \lnot q$.

(a) Then for the proposition $p \rightarrow q =$ “If it snows today, I will ski tomorrow”, the converse is “If I will ski tomorrow, then it snows today”, the contrapositive is “If I will not ski tomorrow, then it doesn’t snow today”, and the inverse is “If it doesn’t snow today, I will not ski tomorrow.”

(b) Then for the propoistion $p \rightarrow q =$ “I come to class whenever there is going to be a quiz” $=$ “If there is going to be a quiz, then I come to class”, then converse is “If I come to class, then there is going to be a quiz”, the contrapositive is “If I don’t come to class, then there is not going to be a quiz”, and the inverse is “If there is not going to be a quiz, then I will not come to class”.

(c) Then for the proposition $p \rightarrow q =$ “A positive integer is a prime only if it has no divisors other than $1$ and itself” $=$ “If a positive integer is a prime, then it has no divisors other than $1$ and itself”, the converse is “If a positive has no divisors other than $1$ and itself, then it is a prime”, the contrapositive is “If a positive integer is not a prime, then it has other divisors than $1$ and itself”, and the inverse is “If a positive integer is not a prime, then it has other divisors than $1$ and itself”.

30 Given the proposition

(a) “If it shows tonight, I will stay home” $=$ “If it snows tonight, then I will stay home” and noting that the definitions in 29, the converse of the proposition is “If I will stay home, then it snows tonight”, the contrapositive is “If I do not stay in home, then it does not snow tonight”, and the inverse is “If it does not snow tonight, I will not stay home”.

(b) “I go to the beach whenever it is a sunny summer day” $=$ “If it is a sunny summer, then I go to the beach”, the converse of the proposition is “If I go to the beach, then it is a sunny summer”, the contrapositive is “If I do not go to the beach, then it is not a sunny summer”, and the inverse is “If it is not a sunny summer, then I won’t go to the beach”

(c) “When I stay up late, it is necessary that I sleep until noon”, whhich is the same as saying “If I sleep until noon, then I stay up late”, the converse of the proposition is “If I stay up late, then I sleep until noon”, the contrapositive is “I won’t stay up late, then I won’t sleep until noon”, and the inverse is “If I don’t sleep until noon, then I won’t stay up late”.

31 Since the number of rows in a truth table is equal to the number of unique propositions squared, that is, the number of rows in a truth table is equal to

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p_1, p_2, \ldots, p_n\})} = 2^n \tag{1} $$

where $\operatorname{ToNum}$ (“Total Number”) is the function mapping the number of unique propositions in a compound proposition in $P$ to some positive integer,

(a) and when the given proposition

$$ P = p \rightarrow \lnot p $$

the proposition has $2^{\operatorname{ToNum}(P)} = 2^1 = 2$ rows.

(b) and when the given proposition

$$ P = (p \lor \lnot r) \land (q \lor \lnot s) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q,r,s\})} = 2^4 = 16 $$

rows.

(c) and when the given proposition

$$ P = q \lor p \lor \lnot s \lor \lnot r \lor \lnot t \lor u $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q,r,s,t,u\})} = 2^6 = 64 $$

rows.

(d) and when the given proposition

$$ P = (p \land r \land t) \leftrightarrow (q \land t) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q,r,t\})} = 2^4 = 16 $$

rows.

32 Using the counting mechanism defined in $(1)$ of 31,

(a) given the proposition

$$ P = (q \rightarrow \lnot p) \lor (\lnot p \rightarrow \lnot q) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q\})} = 2^2 = 4 $$

rows.

(b) given the proposition

$$ P = (p \land \lnot t) \land (p \lor \lnot s) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q,s,t\})} = 2^4 = 16 $$

rows.

(c) given the proposition

$$ P = (p \rightarrow r) \lor (\lnot s \rightarrow \lnot t) \land (\lnot u \rightarrow v) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,r,s,t,u,v\})} = 2^6 = 64 $$

rows.

(d) given the proposition

$$ P = (p \land r \land s) \lor (q \land t) \lor (r \land \lnot t) $$

the proposition has

$$ 2^{\operatorname{ToNum}(P)} = 2^{\operatorname{ToNum}(\{p,q,r,s,t\})} = 2^5 = 32 $$

rows.

33

(a) The truth table for proposition $p \land \lnot p$

(b) The truth table for proposition $p \lor \lnot p$

(c) The truth table for proposition $(p \lor q) \rightarrow q$

(d) The truth table for proposition $(p \lor q) \rightarrow (p \land q)$

(e) The truth table for proposition $(p \rightarrow \lnot p) \leftrightarrow (\lnot q \rightarrow \lnot p)$ is

(f) The truth table for proposition $(p \rightarrow q) \rightarrow (q \rightarrow p)$ is

34

(a)

$$ \begin{array}{ | c | c | c | } \hline p & \lnot p & p \land \lnot p \\ \hline \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{F} \\ \hline \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | } \hline p & \lnot p & p \leftrightarrow \lnot p \\ \hline \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{F} \\ \hline \end{array} $$

(c)

$$ \begin{array}{ | c | c | c | c | } \hline p & q & p \lor q & p \oplus (p \lor q) \\ \hline \text{T} & \text{T} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} \\ \hline \end{array} $$

(d)

$$ \begin{array}{ | c | c | c | c | } \hline p & q & p \land q & p \lor q & (p \land q) \rightarrow (p \lor q) \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \hline \end{array} $$

(e)

(f)

35

(a)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & p \land q & p \oplus q & (p \land q) \rightarrow (p \oplus q) \\ \hline \text{T} & \text{T} & \text{T} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & p \land q & p \oplus q & (p \land q) \rightarrow (p \oplus q) \\ \hline \text{T} & \text{T} & \text{T} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \end{array} $$

(c)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & p \land q & p \lor q & (p \land q) \oplus (p \lor q) \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \end{array} $$

(d)

(e)

(f)

36

(a)

$$ \begin{array}{ | c | c | } \hline p & p \oplus p \\ \hline \text{T} & \text{F} \\ \text{F} & \text{F} \\ \hline \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | } \hline p & \lnot p & p \oplus p \\ \hline \text{T} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} \\ \hline \end{array} $$

(c)

$$ \begin{array}{ | c | c | c | c | } \hline p & q & \lnot q & p \oplus \lnot q \\ \hline \text{T} & \text{T} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \end{array} $$

(d)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & \lnot p & q & \lnot q & \lnot p \oplus \lnot q \\ \hline \text{T} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \hline \end{array} $$

(e)

(f)

37

(a)

$$ \begin{array}{ | c | c | c | c | c | c | } \hline p & q & \lnot q & p \rightarrow \lnot q \\ \hline \text{T} & \text{T} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} \\ hline \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | c | } \hline p & \lnot p & q & \lnot p \leftrightarrow q \\ \hline \text{T} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{F} \\ hline \end{array} $$

(c)

(d)

(e)

38

(a)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & r & p \lor q & (p \lor q) \lor r \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{F} \\ \hline \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & r & p \lor q & (p \lor q) \land r \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{F} \\ \hline \end{array} $$

(c)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & r & p \land q & (p \land q) \lor r \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{F} \\ \hline \end{array} $$

(d)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & r & p \land q & (p \land q) \land r \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{F} \\ \hline \end{array} $$

(e)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & \lnot r & p \lor q & (p \lor q) \land \lnot r \\ \hline \text{T} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{F} \\ \hline \end{array} $$

(f)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & q & \lnot r & p \land q & (p \land q) \lor \lnot r \\ \hline \text{T} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{F} \\ \text{F} & \text{F} & \text{T} & \text{F} & \text{T} \\ \hline \end{array} $$

39

(a)

$$ \begin{array}{ | c | c | c | c | c | } \hline p & \lnot q & r & \lnot q \lor r & p \rightarrow (\lnot q \lor r) \\ \hline \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{F} & \text{F} \\ \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \hline \end{array} $$

(b)

$$ \begin{array}{ | c | c | c | c | c | } \hline \lnot p & q & r & q \rightarrow r & \lnot p \rightarrow (q \rightarrow r) \\ \hline \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{T} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{T} & \text{F} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \end{array} $$

(c)

(d)

(e)

(f)

40

41

42

When either of the two propositions on the either side of the logical or ($\lor$) is true ($\text{T}$), the whole expression is always $\text{T}$. Then in the given expression, $(q \lor \lnot r)$ $\land$ $(r \land \lnot p)$ $\land$ $(r \land \lnot p)$, the compound propositions will always be true. That is if $p$, $q$, and $r$ are all $\text{T}$, then the given proposition is equal to $\text{T}$ $\land$ $\text{T}$ $\land$ $\text{T}$, from where it can be seen that the given proposition is always true.

43 If either one of the propositions $p$, $q$, or $r$ is $\text{T}$ and one if $\text{F}$, in $(p$ $\lor$ $q$ $\lor$ $r)$ $\land$ $(\lnot p$ $\lor$ $\lnot$ $q$ $\lor$ $r)$, then the left hand-side (LHS) of the logical and ($\land$) will always be $T$, no matter what the truth value of the third proposition is.

And on the other hand, reverse is true for the right hand-side (RHS) of the $\land$ connecting the LHS and RHS. That is, the RHS will always be $\text{T}$, one of the propositions, that is, the one being $\text{F}$ will always be $\text{T}$, because of the logical not ($\lnot$).

This means that whenever one of the propositions $p$, $q$, or $r$ is $\text{T}$ and one is $\text{F}$, the LHS and RHS of the given proposition will always be $\T$ and the whole proposition will always be equal to $\text{T}$ $\land$ $\text{T}$. Because of this, the whole expression is then always $\text{T}$.

If $p$, $q$, and $r$ are all $\text{T}$ or $\text{F}$ at the same time, then LHS will be either $\text{T}$ or $\text{F}$ and the RHS the opposite of that, that is, either $\text{F}$ or $\text{T}$. Therefore the given experssion will always be either $\text{T}$ $\land$ $\text{F}$ or vice versa, so the expression will always be $\text{F}$.

44

…

45

…

46

Given that $x = 1$ and assuming that the programming language is short-circuiting w.r.t. it’s boolean logic, the value of $x$ after the expression

(a) $\text{\textbf{if}}$ $x + 2 = 3$ $\text{\textbf{then}}$ $x := x + 1$ means that $x = 2$, because $x + 2$ $=$ $1 + 2$ $=$ $3$ so the $\text{\textbf{then}}$ is run, and so $x$ gets incremented by $1$, i.e. $x$ becomes $2$.

(b) $\text{\textbf{if}}$ $x + 2 = 3$ $\text{\textbf{then}}$ $x := x + 1$ means that $x = 2$, because the left hand-side (LHS) of the $\\operatorname{OR}$ becomes true ($\text{T}$), so no matter what the right hand-side (RHS) is, the $\text{\textbf{then}}$ will be run, and so $x$ gets incremented by $1$, i.e. $x$ becomes $2$.

(c) will be $1$ because the LHS will evaluate to false $(\text{F})$, and in order for the $\operatorname{AND}$ operator to be $\text{T}$, both LHS and RHS needs to be $\text{T}$. This means that the $\text{\textbf{then}}$-part will not be evaluated and $x$ won’t get icremented by the $x := x + 1$ instruction.

(d) $\text{\textbf{if}}$ $(x + 1 = 2)$ $\operatorname{XOR}$ $(x + 2 = 3)$ will evaluate to $\text{T}$ since $x + 1$ $=$ $1 + 1 = 2$ and $x + 2$ $=$ $1 + 3 = 3$, i.e. both propositions of the $\operatorname{XOR}$-operator will evaluate to $\text{T}$ and the whole expression becomes $T$, the $\text{\textbf{then}}$ block will be ran and $x$ gets incremented by one, and will be $x = 2$.

(e) $\text{\textbf{if}}$ $<$ $2$ will evaluate to $T$, since $x = 1 < 2$ is true, and so the $x$ gets incremented by $1$ becoming $2$ after this $\text{\textbf{if}}$-$\text{\textbf{then}}$ instruction is executed.

47

(a)

$$ \begin{array}{ r c c } & 101 & 1110 \\ & 010 & 0001 \\ \hline \operatorname{OR} & 111 & 1111 \\ \operatorname{AND} & 000 & 0000 \\ \operatorname{XOR} & 111 & 1111 \\ \end{array} $$

(b)

$$ \begin{array}{ r c c } & 1111 & 0000 \\ & 1010 & 1010 \\ \hline \operatorname{OR} & 1111 & 1010 \\ \operatorname{AND} & 1010 & 0000 \\ \operatorname{XOR} & 0101 & 1010 \\ \end{array} $$

(c)

$$ \begin{array}{ r c c c } & 00 & 0111 & 0001 \\ & 10 & 0100 & 1000 \\ \hline \operatorname{OR} & 10 & 0111 & 1001 \\ \operatorname{AND} & 00 & 0100 & 0000 \\ \operatorname{XOR} & 10 & 0011 & 1001 \\ \end{array} $$

(d)

$$ \begin{array}{ r c c c } & 11 & 1111 & 1111 \\ & 00 & 0000 & 0000 \\ \hline \operatorname{OR} & 11 & 1111 & 1111 \\ \operatorname{AND} & 00 & 0000 & 0000 \\ \operatorname{XOR} & 11 & 1111 & 1111 \\ \end{array} $$

48

(a)

$$ \begin{array}{ c c c } \\ & 0 & 1011 \\ \lor & 1 & 1011 \\ \hline & 1 & 1011 \\ \land & 1 & 1000 \\ \hline & 1 & 1000 \end{array} $$

(b)

$$ \begin{array}{ c c c } \\ & 0 & 1111 \\ \land & 1 & 0101 \\ \hline & 0 & 0101 \\ \lor & 0 & 1000 \\ \hline & 0 & 1101 \end{array} $$

(c)

$$ \begin{array}{ c c c } \\ & 0 & 1010 \\ \oplus & 1 & 1011 \\ \hline & 1 & 0001 \\ \oplus & 0 & 1000 \\ \hline & 1 & 1001 \end{array} $$

(d)

$$ \begin{array}{ c c c } \\ & 1 & 1011 & & & & 1 & 0001 \\ \lor & 0 & 1010 & & & & 1 & 1011 \\ \hline & 1 & 1011 & & & & 1 & 1011 \\ & & & \searrow & & & & \ \downarrow \\ & & & & 1 & 1011 & & \swarrow \quad \\ \land & & & & 1 & 1011 & \leftarrow & \\ \hline & & & & 1 & 1011 & & \\ \end{array} $$

49

If the truth value of the proposition

$$ \text{P} = \text{Fred is happy} $$

is $0.8$, and the truth value of the proposition

$$ \text{Q} = \text{John is happy}, $$

and the negation of a proposition ($\lnot \text{Q}$) in fuzzy logic is

$$ \phantom{,} \ \lnot \text{Q} = \text{T} - q = 1 - q \ , $$

where $\text{T} = 1$ means true, and $q$ is the (probability) truth value of $\text{Q}$, then

$$ \phantom{.} \ \lnot \text{P} = 1 - 0.8 = 0.2 \ , $$

and

$$ \phantom{.} \ \lnot \text{Q} = 1 - 0.2 = 0.8 \ , $$

50

Since the truth value of the conjunction of two propositions in fuzzy logic is the minimum ($\min$) of the truth values of the two proposition, and given the $\text{P}$ and $\text{Q}$ as in the previous exercise 49, then

Similarly

52

Yes, the assertion “This statement is false” is a proposition. This is because it can be falsified. Let $P$ be the shorthand for the assertion. Then $P$ is just as it was just given, and $\lnot P$ $=$ “This statement is true”.