CS 381 SOLUTIONS TO HOMEWORK 5

Fall 2010

p. 49

46. If A is true then LHS is true. Since $P(x) \vee A$ is true for every x if A is true, RHS is also true.
If A is false then LHS is $\forall x P(x)$. Since $P(x) \vee A$ is P(x) if A is false, RHS is $\forall x P(x)$.
Thus LHS and RHS always take the same value. Hence they are equivalent.

p. 62

38 a) At least one student in this class does not like mathematics.
c) Every student in this class has not taken at least one mathematics course offered at this school.

pp. 73 - 74

14 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

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
$S( movie )$
------------------------
$W( movie )$ -- 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

20 a) No. It is not valid. The second sentence is the converse of the first sentence with a substituted for x and it is not true.

b) Yes, it is valid.

p.85

16. Proof by contradiction, that is we assume that the conclusion (m is even or n is even) is not true and try to derive a contradiction - a statement which is always false.
Suppose the conclusion is not true that is neither m nor n is even.
Since m and n are odd then, there are integers k and l such that m = 2k + 1 and n = 2l + 1.
Then mn = (2k+1)(2l+1) = 4kl + 2k + 2l + 1.
Since 4kl, 2k and 2l are even, 4kl+2k+2l+1 is odd, that is mn is odd which contradicts the hypothesis that mn is even.
Thus our assumption is wrong and either m or n must be even.