CS 381 Solutions to Homework 5

pp. 36 - 37

26 (b) $\neg \exists x P(x)$, where $P(x)$ means x has lost more than one thousand dollars playing the lottery.
$\exists x P(x)$
Someone has lost more than one thousand dollars playing the lottery.

(d) $\neg \exists x P(x)$, where $P(x)$ means x has sent e-mail to exactly two studnets in the class.
$\exists x P(x)$
Some student in the class has sent e-mail to exactly two other studnets in the class.

(f) $\neg \exists x P(x)$, where $P(x)$ means x has solved at least one exercise in every section of this book.
$\exists x P(x)$
Some student has solved at least one exercise in every section of this book.

38. These do not hold, because if $A$ contains $x$ in it such as when $A$ is $Q(x)$, then it is not true. For example let $P(x)$ represent $x$ is happy and let $Q(x)$ represent $x$ is rich and see how they read.

This question should read "... A is a proposition not involving the variable $x$".

pp. 183 - 185

10 a) $\forall x [ R(x) \rightarrow T(x) ]$
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$R(Linda) \rightarrow T(Linda)$ -- Universal Instantiation
$R(Linda)$
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$T(Linda)$ -- Modus Ponens
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$\exists x T(x)$ -- Existential Generalization

b) Apply Universal Instantiation to $\forall x [ D(x) \rightarrow A(x) ]$ for each of the five, then Modus Ponens to each to obtain $A( Melissa )$, $A( Aaron )$, $A( Ralph )$, $A( Veneesha )$, and $A( Keeshawn )$.

c) $\exists x [S(x) \wedge C(x) ]$
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$S( movie ) \wedge C( movie )$ -- Existential Instantiation
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$S( movie )$ -- Simplification

$\forall x [ S(x) \rightarrow W(x) ]$
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$S( movie ) \rightarrow W( movie )$ -- Universal Instantiation
$S( movie )$
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$W( movie )$ -- Modus Ponens

Also
$S( movie ) \wedge C( movie )$
------------------------
$C( movie )$ -- Simplification

Hence
$S( movie ) \wedge C( movie )$
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$\exists x [W(x) \wedge C(x) ]$ -- Existential Generalization

d) $\exists x [F(x) \wedge C(x) ]$
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$F(c) \wedge C(c)$ -- Existential Instantiation
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$F(c)$ -- Simplification

$\forall x [ F(x) \rightarrow L(x) ]$
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$F(c) \rightarrow L(c)$ -- Universal Instantiation
$F(c)$
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$L(c)$ -- Modus Ponens

Also
$F(c) \wedge C(c)$
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$C(c)$ -- Simplification

$L(c)$
$C(c)$
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$L(c) \wedge C(c)$ -- Conjunction
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$\exists x [ C(x) \wedge L(x) ]$ -- Existential Generalization

12. "n does not equal 3l for some integer l" in the second sentence is what is to be proven: "n is not divisible by 3". That makes this "proof" circular.

16 (c) Suppose that the square of an even number is not even. Let $n$ be an even number. Then there is an integer $k$ that satisfies $n = 2k$. Hence $n^{2} = 4k^{2}$. Since $4k^{2}$ is divisible by 2, if $n^{2}$ is odd, then that means that an odd number is divisible by 2. That is a contradiction. Hence the square of an even number must be an even number.