Answer
a. All dogs cannot learn new tricks.
b. There is a rabbit that knows calculus.
c. There is a bird that cannot fly.
d. There is a dog that can talk.
e. There is someone in this class who knows French and Russian.
Work Step by Step
a) Let the domain be dogs and P(x) mean "x can learn new tricks“. We can then rewrite the given statement as:
$\exists xP(x)$
The negation is then by De Morgan’s Law for Qualifiers:
$\neg (\exists xP(x)) \equiv \forall x \neg P(x)$
This means: All dogs cannot learn new tricks.
b) Let the domain be rabbits and Q(x) mean "x knows calculus”. We can then rewrite the given statement as:
$\forall x \neg Q(x)$
The negation is then by De Morgan’s Law for Qualifiers and the double negation law:
$\neg (\forall x \neg Q(x)) \equiv \exists x \neg (\neg Q(x)) \equiv \exists x Q(x) $
This means: There is a rabbit that knows calculus.
c) Let the domain be birds and R(x) mean ”x can fly”. We the given statement as:
$\forall xR(x)$
The negation is then by De Morgan’s Law for Qualifiers:
$\neg (\forall xR(x)) \equiv \exists x \neg R(x)$
This means: There is a bird that cannot fly.
d) Let the domain be dogs and S(x) mean " x can talk". We can then rewrite the given statement as "every monkey can not speak French":
$\neg \exists x S(x)$
The negation is then by the double negation law:
$\neg (\neg \exists x S(x)) \equiv \exists x S(x)$
This means: There is a dog that can talk.
e) Let the domain be the people in this class and T(x) mean "x knows French" and U(x) mean "x knows Russian". We can then rewrite the given statement as:
$\neg \exists x(T(x) \land U(x))$
The negation is then by the double negation law:
$\neg(\neg \exists x(T(x) \land U(x))) \equiv \exists x(T(x) \land U(x)) $
This means: There is someone in this class who knows French and Russian.
