Suppose that \(\alpha\) is a wff not containing the negation symbol \(\lnot\)

Claim: The length of \(\alpha\) is odd

Throughout the problem I use \(\square\) to denote an arbitrary choice of \(\land,\lor,\to,\leftrightarrow\)

Let \(\langle \varepsilon_1,\ldots,\varepsilon_n\rangle\) be its corresponding construction sequence. If \(n=1,2\) then the claim is obvious, since \(\alpha\) not containing the \(\lnot\) symbol implies that \(\varepsilon_1\) and \(\varepsilon_2\) are sentence symbols. Assume \(n > 2\). I prove the claim inductively. If \(\varepsilon_3\) is a sentence symbol, then certainly \(\varepsilon_3\) has odd length. If \(\varepsilon_3 = \varepsilon_\square(\varepsilon_1,\varepsilon_2)\), then \(\varepsilon_3 = (\varepsilon_1\square\varepsilon_2)\) has odd length since \(\varepsilon_1\) and \(\varepsilon_2\) have odd length. Thus, \(\varepsilon_3\) has odd length. Now, assume \(\varepsilon_1,\ldots,\varepsilon_k\) have odd length. If \(\varepsilon_{k+1}\) is a sentence symbol then it is obviously of odd length. On the other hand, if \(\varepsilon_{k+1} = \varepsilon_\square(\varepsilon_i,\varepsilon_j)\) where \(i,j < k+1\), then \(\varepsilon_{k+1} = (\varepsilon_i\square\varepsilon_j)\) has odd length. Therefore \(\varepsilon_{k+1}\) has odd length. From this argument it follows that the last construction step of \(\alpha\), that is \(\varepsilon_n\), has odd length, and thus \(\alpha\) has odd length.

Claim: More than a quarter of the symbols in \(\alpha\) are sentence symbols.

To show this, I will first show that the length of \(\alpha\) is of the form \(4k+1\) for some integer \(k\), and that \(\alpha\) contains \(k+1\) sentence symbols.

As before, \(\varepsilon_1\) and \(\varepsilon_2\) must be sentence symbols, so they have length \(1\). Since they contain \(1\) sentence symbols, it’s clear that \(\varepsilon_1\) and \(\varepsilon_2\) have lengths of the form \(4k+1\) and contain \(k+1\) sentence symbols, where \(k=0\). Now, assume that \(\varepsilon_1,\ldots,\varepsilon_j\) have lengths of the form \(4k+1\) and contain \(k+1\) sentence symbols. If \(\varepsilon_{j+1}\) is a sentence symbol then we already know it is also of this form. On the other hand, consider the case where \(\varepsilon_{j+1} = \varepsilon_\square(\varepsilon_r,\varepsilon_s)\) for some \(r,s < j+1\). Assume that \(\varepsilon_r\) has length \(4q + 1\) for some integer \(q\), and thus contains \(q+1\) sentence symbols. Assume that \(\varepsilon_s\) has length \(4p+1\) for some integer \(p\), and thus contains \(p+1\) sentence symbols. Notice that \(\varepsilon_{j+1} = (\varepsilon_r\square\varepsilon_s)\) then contains \(q + p + 2\) sentence symbols, and is of length \(4q + 1 + 4p + 1 + 3 = 4(q +p+1) + 1\). We notice that \(\varepsilon_{j+1}\) has length of the form \(4k + 1\) and \(k+1\) sentence symbols where \(k = q + p + 1\). By induction, this argument yields that \(\varepsilon_n = \alpha\) has length of the form \(4k+1\) and \(k+1\) sentence symbols. Since \(k+1 > \frac{4k+1}{4}\), it’s clear that more than a quarter of the symbols in \(\alpha\) are sentence symbols.$$