Answer
a. There is a dog that does not have fleas.
b. All horses can not add.
c. There is a koala that cannot climb.
d. A monkey can speak French.
e. All pigs can not swim or not catch fish.
Work Step by Step
a) a) Let the domain be dogs and P(x) mean "x has fleas”. We can then rewrite the given statement as:
$\forall xP(x)$
The negation is then by De Morgan’s Law for Qualifiers:
$\neg (\forall xP(x)) \equiv \exists x\neg P(x)$
This means: There is a dog that does not have fleas.
b) Let the domain be horses and Q(x) mean "x can add”. We can then rewrite the given statement as:
$\exists xQ(x)$
The negation is then by De Morgan’s Law for Qualifiers:
$\neg (\exists xQ(x)) \equiv \forall x \neg Q(x)$
This means: All horses can not add.
c) c) Let the domain he koalas and R(x) mean “x can climb”, rewrite the given statement as:
$\forall x R(x)$
The negation is then by De Morgan’s Law for Qualifiers:
$\neg (\forall x R(x)) \equiv \exists x \neg R(x)$
This means: There is a koala that cannot climb.
d) Let the domain he monkeys and S(x) mean ’ x can speak French' . We can then rewrite the given statement as ”every monkey can not. speak French":
$\forall x(\neg S(x))$
The negation is then by De Morgan’s Law for Qualifiers and the double negation law:
$\neg (\forall x(\neg S(x))) \equiv \exists x \neg (\neg S(x)) \equiv \exists x S(x) $
This means: A monkey can speak French.
e) Let the domain be pigs and T(x) mean "x can swim" and U(x) mean "x can catch fish“. We can then rewrite the given statement as:
$\exists x(T(x) \land U(x))$
The negation is then by De Morgan’s Law for Qualifiers and the regular De Morgan’s Law:
$\neg (\exists x(T(x) \land U(x))) \equiv \forall x \neg(T(x) \land U(x)) \equiv \forall x (\neg T(x) \lor \neg U(x)) $
This means: All pigs can not swim or not catch fish.
