Question

Express each of these statements using quantifiers. Then form the negation of the statement, so that no negation is to the left of a quantifier. Next, express the negation in simple English. (Do not simply use the phrase “It is not the case that.”)

a) Some old dogs can learn new tricks.

b) No rabbit knows calculus.

c) Every bird can fly.

d) There is no dog that can talk.

e) There is no one in this class who knows French and Russian.

Short answer

a) The statement "Some old dogs can learn new tricks" expressed using quantifiers would be$\mathbf{\neg }\mathbf{(}\exists \mathbf{xP}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$; and the negation in simple English would be " All dogs cannot learn new tricks".

b) The statement “No rabbit knows calculus “expressed using quantifiers would be$\mathbf{\neg }\mathbf{(}\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{Q}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{Q}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{xQ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$; and the negation in simple English would be " There is a rabbit that knows calculus ".

c)The statement “Every bird can fly “expressed using quantifiers would be $\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathbf{X}}_{\mathrm{i}}^2$; and the negation in simple English would be " There is a bird that cannot fly ".

d) The statement "There is no dog that can talk" expressed using quantifiers would be $\mathbf{\neg }\mathbf{(}\mathbf{\forall }\mathbf{x}\mathbf{(}\mathbf{\neg }\mathbf{S}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{S}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{xS}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$; and the negation in simple English would be " There is a dog that can talk

e) The statement "There is no one in this class who knows French and Russian" expressed using quantifiers would be$\mathbf{\neg }\mathbf{(}\mathbf{\neg }\exists \mathbf{x}\mathbf{(}\mathbf{T}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\wedge }\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\left(}\mathbf{T}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{U}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}$; and the negation in simple English would be" There is some in this class who knows French and Russian ".

Step-by-step solution

INTERPRETATION SYMBOLS

Negation$\mathbf{\neg }\mathbf{p}$: not p

Conjunction $\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{:}\mathbf{p}$and q

Existential quantification $\exists \mathbf{xP}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$: There exists an element xin the domain such that P(x).

Universal quantification $\mathbf{\forall }\mathbf{xP}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{:}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$for all values of x in the domain.

LOGICAL EQUIVALENCES

Double negation law:

$\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\equiv }\mathbf{p}$

De Morgan's laws:

$$\begin{gathered} \mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\equiv }\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{q} \\ \mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\equiv }\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{\neg }\mathbf{q} \end{gathered}$$

De Morgan's Laws for Qualifiers:

$$\begin{gathered} \mathbf{\neg }\exists \mathbf{xp}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\equiv }\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{\neg }\mathbf{\forall }\mathbf{xP}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

¬(∀xP(x))≡∃x¬P(x)

a) Let the domain be dogs and P(x)mean "x can learn new tricks". We can then rewrite the given statement as:

$\exists \mathbf{xP}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

The negation is then by De Morgan's Law for Qualifiers:

$\mathbf{\neg }\mathbf{(}\exists \mathbf{xP}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

This means: All dogs cannot learn new tricks.

b) Let the domain be rabbits and Q(x)mean "x knows calculus". "No rabbit ..." can be rewritten as "All rabbits do not...".

We can then rewrite the given statement as:

role="math" localid="1668581308838" $\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

The negation is then by De Morgan's Law for Qualifiers and the double negation law:

$\mathbf{\neg }\mathbf{(}\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{Q}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{Q}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{xQ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

This means: There is a rabbit that knows calculus.

∀xR(x)

c) Let the domain be birds and R(x) mean "x can fly". We can then rewrite the given statement as:

$\mathbf{\forall }\mathbf{xR}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

The negation is then by De Morgan's Law for Qualifiers:

$\mathbf{\neg }\mathbf{(}\mathbf{\forall }\mathbf{xR}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{R}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

This means: There is a bird that cannot fly.

∀x(¬S(x))

d) Let the domain be dogs and S(x)mean "x can talk". We can then rewrite the given statement as "every monkey cannot speak French":

$\mathbf{\neg }\exists \mathbf{xS}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

The negation is then by the double negation law:

$\mathbf{\neg }\mathbf{(}\mathbf{\forall }\mathbf{x}\mathbf{(}\mathbf{\neg }\mathbf{S}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{S}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{xS}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

This means: There is a dog that can talk.

¬∃x(T(x)∧U(x))

e) Let the domain be the people in this class and T(x)mean "xknows French" and U(x)mean "x knows Russian".

We can then rewrite the given statement as:

$\mathbf{\neg }\exists \mathbf{x}\mathbf{\left(}\mathbf{T}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{U}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}$

The negation is then by the double negation law:

$\mathbf{\neg }\mathbf{(}\mathbf{\neg }\exists \mathbf{x}\mathbf{(}\mathbf{T}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\wedge }\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left)}\mathbf{\right)}\mathbf{\equiv }\exists \mathbf{x}\mathbf{\left(}\mathbf{T}\mathbf{\right(}\mathbf{x}\mathbf{)}\mathbf{\wedge }\mathbf{U}\mathbf{(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}$

This means: There is someone in this class who knows French and Russian.