Page 74 - HOW TO PROVE IT: A Structured Approach, Second Edition

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60 Quantificational Logic
(ii) ∀x[∃y(y is a friend of x and y has the measles) → x will have to be
quarantined].
Now, letting F(y, x) stand for “y is a friend of x,” M(y) for “y has the
measles,” and Q(x) for “x will have to be quarantined,” we get:
(iii) ∀x[∃y(F(y, x) ∧ M(y)) → Q(x)].
4. The word anyone is difficult to interpret, because in different statements it
means different things. In statement 3 it meant everyone, but in this statement
it means someone. Here are the steps of our analysis:
(i) (Someone in the dorm has a friend who has the measles) → (everyone
in the dorm will have to be quarantined).
(ii) ∃x(x lives in the dorm and x has a friend who has the measles) →∀z(if
z lives in the dorm then z will have to be quarantined).
Using the same abbreviations as in the last statement and letting D(x) stand
for “x lives in the dorm,” we end up with the following formula:
(iii) ∃x[D(x) ∧∃y(F(y, x) ∧ M(y))] →∀z(D(z) → Q(z)).
5. Clearlytheanswerwillhavetheformofaconditionalstatement,(A ⊆ B) →
(A and C \ B are disjoint). We have already written A ⊆ B symbolically in
Example 2.1.2. To say that A and C \ B are disjoint means that they have no
elements in common, or in other words ¬∃x(x ∈ A ∧ x ∈ C \ B). Putting
this all together, and filling in the definition of C \ B, we end up with
∀x(x ∈ A → x ∈ B) →¬∃x(x ∈ A ∧ x ∈ C ∧ x /∈ B).

When a statement contains more than one quantifier it is sometimes difficult
to figure out what it means and whether it is true or false. It may be best in this
case to think about the quantifiers one at a time, in order. For example, consider
the statement ∀x∃y(x + y = 5), where the universe of discourse is the set of
all real numbers. Thinking first about just the first quantifier expression ∀x,
we see that the statement means that for every real number x, the statement
∃y(x + y = 5) is true. We can worry later about what ∃y(x + y = 5) means;
thinking about two quantifiers at once is too confusing.
If we want to figure out whether or not the statement ∃y(x + y = 5) is true for
every value of x, it might help to try out a few values of x. For example, suppose
x = 2. Then we must determine whether or not the statement ∃y(2 + y = 5) is
true. Now it’s time to think about the next quantifier, ∃y. This statement says
that there is at least one value of y for which the equation 2 + y = 5 holds. In
other words, the equation 2 + y = 5 has at least one solution. Of course, this is
true, because the equation has the solution y = 5 − 2 = 3. Thus, the statement
∃y(2 + y = 5) is true.
Let’s try one more value of x.If x = 7, then we are interested in the statement
∃y(7 + y = 5),whichsaysthattheequation7 + y = 5hasatleastonesolution.