Section 1.3 Quantifiers, Predicates, and Validity
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whenever the antecedent is true, so is the consequent, and the implication is therefore true.
b. The wff
(4x)P(x) S P(a)
is valid because in any interpretation, a is a particular member of the domain and
therefore has the property that is shared by all members of the domain.
c. The wff
(4x)[P(x) ` Q(x)] 4 (4x)P(x) ` (4x)Q(x)
is valid. If both P and Q are true for all the elements of the domain, then P is true
for all elements and Q is true for all elements, and vice versa.
d. The wff
P(x) S [Q(x) S P(x)]
is valid, even though it contains a free variable. To see this, consider any interpretation, and let x be any member of the domain. Then x either does or does not
have property P. If x does not have property P, then P(x) is false; because P(x)
is the antecedent of the main implication, this implication is true. If x does have
property P, then P(x) is true; regardless of the truth value of Q(x), the implication
Q (x) S P(x) is true, and so the main implication is also true.
e. The wff
(E x)P(x) S (4x)P(x)
is not valid. For example, in the interpretation where the domain consists of the
integers and P(x) means that x is even, it is true that there exists an integer that is
even, but it is false that every integer is even. The antecedent of the implication is
true and the consequent is false, so the value of the implication is false.
We do not necessarily have to go to a mathematical context to construct an
interpretation in which a wff is false, but it is frequently easier to do so because
the relationships among objects are relatively clear.
pRaCtiCe 21 Is the wff valid or invalid? Explain.
(4x)[P(x) ~ Q(x)] S (4x)P(x) ~ (4x)Q(x)