# Section 2.4: Problem 1 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

For a term
, let
be the expression obtained from
by replacing the variable
by the term
. Restate this definition without using any form of the word “replace” or its synonyms.

*Suggestion*: Use recursion on . (Observe that from the new definition it is clear that is itself a term.)
As suggested, we define
using recursion. Let
be the set of variables and constant symbols,
be the set of terms, and
be defined

Then, since
is freely generated from
by
where
is an
-place function symbol and
, there exists a unique extension
such that a)
if
, and b)
. We define
to be
.

Now, once
is defined, we can refer to the forbidden word “replace” to verify that
is indeed what we intended it to be. For
, we have
, which is
with
replaced by
. For any other variable symbol
, we have
, and for any constant symbol
, we have
, that is, in both cases, the term is left unchanged as there is no
in it at all. For any other term of the form
, we have
, i.e.
is obtained by substituting
for
in each term
,
.

Finally, we observe that by definition, the extension
, i.e.
is a term.