# Section 2.5: Problem 4 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let
. Is
consistent? Is
satisfiable?

According to the soundness and completeness theorems, the answers to both questions should be the same. It is easy to see that
is satisfiable. Consider
where
and
such that
. Then, certainly,
,
and
, hence,
.

Note. At first, it may seem to some that
is neither consistent nor satisfiable. Indeed, the first formula says that
does not hold for every element in the domain, and then it lists all the possible variables, saying that
holds for all of them. The problem with this approach is that what the first formula says is that, indeed, there is an element in the domain for which
does not hold. However, after that,
does not list all the elements of the domain, rather, it contains an infinite list of independent variables saying that whatever element is assigned to each of them,
holds for the element. In other words, the rest of
just says that there is an element for which
holds. In other words,
is somewhat (not really logically) equivalent to
, in the sense that for every domain for which
can be satisfied with some
,
(consisting of sentences only) is satisfied, and vice versa.