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