Page 83 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Equivalences Involving Quantifiers 69
that ∀x ∈ AP(x) means the same thing as ∀x(x ∈ A → P(x)). For exam-
2
−
ple, the formula ∀x ∈ R ∃y ∈ R (y = x) discussed earlier means the same
+
2
thing as ∀x(x ∈ R →∃y ∈ R (y = x)), which in turn can be expanded as
−
+
2
∀x(x ∈ R →∃y(y ∈ R ∧ y = x)). By the definitions of R and R ,an
+
−
−
+
2
equivalent way to say this would be ∀x(x > 0 →∃y(y < 0 ∧ y = x)). You
should make sure you are convinced that this formula, like the original for-
mula, means that every positive real number has a negative square root. For
another example, note that the statement A ⊆ B, which by definition means
∀x(x ∈ A → x ∈ B), could also be written as ∀x ∈ A(x ∈ B).
It is interesting to note that the quantifier negation laws work for bounded
quantifiers as well. In fact, we can derive these bounded quantifier negation laws
from the original laws by thinking of the bounded quantifiers as abbreviations,
as described earlier. For example,
¬∀x ∈ AP(x)
is equivalent to ¬∀x(x ∈ A → P(x)) (expanding abbreviation),
which is equivalent to ∃x¬(x ∈ A → P(x)) (quantifier negation law),
which is equivalent to ∃x¬(x /∈ A ∨ P(x)) (conditional law),
which is equivalent to ∃x(x ∈ A ∧¬P(x)) (DeMorgan’s law),
which is equivalent to ∃x ∈ A¬P(x) (abbreviation).

Thus, we have shown that ¬∀x ∈ AP(x) is equivalent to ∃x ∈ A¬P(x). You
are asked in exercise 5 to prove the other bounded quantifier negation law, that
¬∃x ∈ AP(x) is equivalent to ∀x ∈ A¬P(x).
It should be clear that if A =∅ then ∃x ∈ AP(x) will be false no matter
what the statement P(x) is. There can be nothing in A that, when plugged in
for x, makes P(x) come out true, because there is nothing in A at all! It may
not be so clear whether ∀x ∈ AP(x) should be considered true or false, but we
can find the answer using the quantifier negation laws:

∀x ∈ AP(x)
is equivalent to ¬¬∀x ∈ AP(x) (double negation law),
which is equivalent to ¬∃x ∈ A¬P(x) (quantifier negation law).

Now if A = ∅ then this last formula will be true, no matter what the statement
P(x) is, because, as we have seen, ∃x ∈ A¬P(x) must be false. Thus, ∀x ∈
AP(x) is always true if A = ∅. Mathematicians sometimes say that such a
statement is vacuously true. Another way to see this is to rewrite the statement
∀x ∈ AP(x) in the equivalent form ∀x(x ∈ A → P(x)). Now according to the
truth table for the conditional connective, the only way this can be false is if
there is some value of x such that x ∈ A is true but P(x) is false. But there is