# Section 2.4: Problem 13 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

Show that
.

*Remarks*: More generally, if is substitutable for in and does not occur in , then Thus the formula offers an alternative of sorts to the substitution .

We need to show that
and
.

For the first formula, by the deduction and generalization theorems, it is sufficient to show that
. By MP, it is sufficient to show that
. Now, by axiom group 6, we have
. By applying MP twice, we obtain
, and by using the deduction theorem twice, we get
.

The second formula, by Rule T, follows from
and
, where the second one belongs to axiom group 2. Now, for the first one, by axiom group 3, MP and generalization theorem, it is sufficient to show that
. Now,
belongs to axiom group 6,
is a tautology (axiom group 1), and, by Rule T, we obtain
.