# Section 1.5: Problem 11 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

(Interpolation theorem) If
, then there is some
all of whose sentence symbols occur both in
and in
and such that
.

*Suggestion*: Use the preceding exercise.*Remarks*: There is an analogue of Exercise 11 that holds for first-order logic. But the proof in that case is very different, because there is no analogue of Exercise 10.

Consider all sentence symbols
that are used in
but not used in
. Define
, and
,
. Then, according to Exercise 10,
, and since no sentence symbol
is used in
,
implies
. Therefore,
and
uses only those sentence symbols in
that are also in
.

There might still be some sentence symbols in
that are not in
, and, hence, not in
, but we don’t care. For example,
, therefore,
, where
is tautologically equivalent to
constructed.