Calculating the number of variables and constants in a term

46 Views Asked by At

I'm reading Kees Doets's Basic Model Theory and couldn't get around the first exercise, which is rather simple (no doubt my lack of arithmetical skills played a part). Let $t$ be a term and let $n_i$ ($i = 1, 2, 3, \dots$) be the number of $i$-ary function symbols that occur in $t$. The exercise is to provem by term induction, that the number of variables and constants in $t$ is $1+\sum_i(i-1) n_i$, that is, $1 + n_2 + 2n_3 + \dots$.

The base case is obvious. As for the induction step, suppose $t$ is $f(t_1, \dots, t_k)$. The induction hypothesis hold for each $t_l (1 \leq l \leq k)$, so we know that, for each such $t_l$, the above formula holds. Further, we know that the number of variables and constants in $t$ is the sum of the number of constants and variables in each $t_l$ (right?). Here comes my problem: doesn't that result in the number $k + \sum_i(i-1)n_i$? Where did I go wrong? It's probably something very simple, but I couldn't figure out what, exactly, is the problem.

1

There are 1 best solutions below

4
On BEST ANSWER

Ihe number of occurrences of $k$-ary function symbols in $f(t_1,\dots,t_k)$ is $1$ more than the sum of the numbers of occurrences of $k$-ary function symbols in the $t_i$.

More formally, if $n_k$ is the combined number of occurrences of $k$-ary function symbols in the $t_i$, then the number $n'_k$ of occurrences of $k$-ary function symbols in $f(t_1,\dots,t_k)$ is given by $n'_k=1+n_k$.

In the formula, the summand $(k-1)n_k$ becomes $(k-1)n'_k$, that is, $(k-1)+(k-1)n_k$.

That yields an additional $k-1$ in the sum, and your $k$ in front is $1+(k-1)$.