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Homework 3 Solution
p. 46 E1: "Alice and Bob are both spies."
Symbolization key:

$A$ : "Alice is a spy;

$B$ : "Bob is a spy
Sentence:

$A \wedge B$

p.46 E4: "The Germany embassy will be in an uproar, unless someone has broken the code."
Symbolization key:

$G$ : The German embassy will be in an uproar;

$C$ : Someone has broken the code.

p.46 E6: "Either Alice or Bob is a spy, but not both."
Symbolization key:

$A$ : "Alice is a spy.";

$B$ : "Bob is a spy."
Sentence:

$(A \vee B) \wedge \neg (A \wedge B)$

p. 47 H2: "If Doctor Octopus gets the uranium, he will blackmail the city. I am certain of this beacuse if Doctor Octopus gets the uranium, he can make a dirty bomb, and if he can make a dirty bomb, he will blackmail the city."
Symbolization key:

$U$ : "Doctor Octopous gets the uranium";

$C$ : "Doctor Octopus will blackmail the city.";

$B$ : "Doctor Octopus can make a dirty bomb."
Argument:

$(U \rightarrow B) \wedge (B \rightarrow C) \therefore (U \rightarrow C)$

p. 55 C: The scope of the first

$\rightarrow$ is

$H \rightarrow I$ . The scope of the first

$\vee$ is

$(H \rightarrow I) \vee (I \rightarrow H)$ . The scope of the second

$\rightarrow$ is

$I \rightarrow H$ . The scope of the

$\wedge$ is

$[(H \rightarrow I) \vee (I \rightarrow H)] \wedge (J \vee K)$ . The scope of the second

$\vee$ is

$J \vee K$ .

p. 76 C1: Write a complete truth table for

$\neg(S \leftrightarrow (P \rightarrow S))$

$\begin{array}{l|l|lll} S&P&\neg(S & \leftrightarrow (P & \rightarrow S)) \\ \hline T & T & \mathbf{F} & T & T \\ T & F & \mathbf{F} & T & T \\ F & T & \mathbf{F} & T & F \\ F & F & \mathbf{T} & F & T \\ \end{array}$ p. 76 D4: Write a complete truth table for

$[(D \wedge R) \rightarrow I] \rightarrow \neg (D \vee R)$

$\begin{array}{l|l|l|llllll} D&R&I& [(D &\wedge R)&\rightarrow I] &\rightarrow &\neg(D &\vee R) \\ \hline T & T & T & & T & T & \mathbf{F} & F & T \\ T & T & F & & T & F & \mathbf{T} & F & T \\ T & F & T & & F & T & \mathbf{F} & F & T \\ T & F & F & & F & T & \mathbf{F} & F & T \\ F & T & T & & F & T & \mathbf{F} & F & T \\ F & T & F & & F & T & \mathbf{F} & F & T \\ F & F & T & & F & T & \mathbf{T} & T & F \\ F & F & F & & F & T & \mathbf{T} & T & F \\ \end{array}$