If $\epsilon$ is a term, then $K(\epsilon)=1$ and every terminal segment $\epsilon'$ of $\epsilon$ is a concatenation of one or more terms (Lemma 23B), therefore, $K(\epsilon')>0$ (equals the number of terms in the concatenation).

We now prove the stronger result, as suggested. We also prove the uniqueness of the decomposition of such $\epsilon$ into $K(\epsilon)$ terms. We prove it by induction on the length of $\epsilon$ .

For $\epsilon$ consisting of $1$ symbol, $\epsilon$ cannot be a function symbol, as otherwise $K(\epsilon)\le0$ . Hence, $\epsilon$ is either a variable or constant symbol. In both cases $\epsilon$ is a term, and $K(\epsilon)=1$ .

Now, suppose that for all lengths less than $l$ the statement is true, and we consider $\epsilon$ of length $l$ . Its terminal segment $\epsilon'$ of length $l-1$ satisfies the condition that every terminal segment $\epsilon''$ of $\epsilon'$ is a terminal segment of $\epsilon$ , hence, $K(\epsilon'')>0$ . Therefore, $\epsilon'$ is a (unique) concatenation of $m=K(\epsilon')>0$ terms. If the first symbol of $\epsilon$ is a variable or constant symbol, then $K(\epsilon)=m+1$ . The first variable or constant symbol of $\epsilon$ is a term itself. And no term of length greater than 1 can start with a variable or constant (by definition). Moreover, $\epsilon'$ is uniquely decomposed into $m$ terms, therefore, $\epsilon$ is uniquely decomposed into $m+1=K(\epsilon)$ terms. If the first symbol of $\epsilon$ is an $n$ -place function symbol $f$ , then, since $K(\epsilon)=m+1-n>0$ , $n\le m$ . Then, $f$ together with the first $n$ terms of the unique decomposition of $\epsilon'$ into $m\ge n$ terms forms one new term, and the resulting number of terms is $m-n+1$ , as instead of $n$ terms we now have just $1$ term. So, again, the number of terms in the decomposition is $m-n+1=K(\epsilon)$ . The composition is also unique, as, by definition, a term starting with $f$ must be $ft_{1}\ldots t_{n}$ , where $t_{1},\ldots,t_{n}$ are terms, and due to the unique decomposition of $\epsilon'$ into terms, $t_{1},\ldots,t_{n}$ are uniquely determined, and so is the decomposition of the remaining expression into $m-n$ terms.