# Section 1.1: Problem 3 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 $\alpha$ be a wff; let $c$ be the number of places at which binary connective symbols ($\wedge$ , $\vee$ , $\rightarrow$ , $\leftrightarrow$ ) occur in $\alpha$ ; let $s$ be the number of places at which sentence symbols occur in $\alpha$ . (For example, if $\alpha$ is $(A\rightarrow(\neg A))$ then $c=1$ and $s=2$ .) Show by using the induction principle that $s=c+1$ .
A single sentence symbol wff has $s=1$ sentence symbol and $c=0$ binary connective symbols. Now, if $\alpha$ and $\beta$ are two wffs each having sentence symbols ($n$ and $m$ , respectively) one more than binary connective symbols, the negation of $\alpha$ does not change the number of either sentence symbols or binary connective symbols, and any wff $(\alpha\star\beta)$ , where $\star$ is one of the binary connective symbols, has $n+m$ sentence symbols and $n-1+1+m-1=n+m-1$ binary connective symbols. Using the induction principle outlined in the textbook, we are done.