# Section 1.4: Problem 2 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
Obviously $(A_{3}\rightarrow\wedge A_{4})$ is not a wff. But prove that it is not a wff.
It has 6 symbols (see Exercise 2, Section 1.1). :) Actually, we can put it more formally. The function defined in the example on p.43 specifies the length of any wff. If $\overline{h}(\alpha)=6$ , then either and $\overline{h}(\beta)=3$ , or $\alpha=(\gamma\wedge\delta)$ and $\overline{h}(\gamma)+\overline{h}(\delta)=3$ . At the same time, the formulas for $\overline{h}$ ensure that each value of $\overline{h}$ is either $1$ or at least $4$ . Hence, neither case is possible.
Another way to prove the statement, is to prove by induction that the combination of symbols $\rightarrow\wedge$ is not possible in any wff. This is easily shown by induction together with the statement that no wff can start or end with a binary connective symbol.
Yet another way to show this is to prove that in any wff the number of pairs of parentheses equals the number of connective symbols. This is true for sentence symbols, and each building operation adds exactly one connective symbol and one pair of parentheses. See also Exercise 4, Section 1.3.