The cardinality of a language is defined to be the set of expressions over the set of variables $V$, the set of functions $F$ and the set of predicate symbol $P$. now I want to decide for any finite set of variables $V$, how can we get some conclusion about the set of expressions in the language? Under what condition, the set of the expressions is of the same cardinality as the set of variables?
I think for the $V=\emptyset$, for the set of expression is empty, we require there are not any $0$-ary function symbol and predicate symbol. For $|V|=1$, for the set of expressions to have cardinality $1$, we need either:
1.only one nullary predicate symbol, no function symbol.
2.No function symbol, no nullary predicate symbol. Only 1 unary predicte symbol.
And for $|V|=2,3,4...$, it is even more complicated. So for finitely many variables, when can we have the set of wff(expressions) have the same cardinality to $V$?
Thanks for any help.
As long as $V \neq \varnothing$, if there is a non-nullary function symbol, then there are infinitely many terms. Suppose $V$ finite. Next, let $n$ be the number of nullary function symbols and for any $k$, $n_k$ the number of $k$-ary predicates. Let $a$ be the maximal arity of your predicates. Let $m=n+|V|$. In the following we suppose $|V|>1$ (since you already did the first cases). Then you have $n_0+mn_1+\dots+m^an_a$ expressions (for any $k$-ary predicate, there are $m^k$ possible tuples of arguments). So either $n>0$ and the only possibility is $n_0=|V|$ and $n_k=0$ for $k \neq 0$ (but in this case you can have as many non-nullary symbol functions as you want since the expressions countains no terms). Or $n=0$ and the only two possibilities are $n_0=|V|$ and $n_k=0$ for $k \neq 0$ (and then again, as many non-nullary functions as you want) or $n_1=1$ and $n_k=0$ for $k \neq 1$. Of course I suppose you use no logical connectors. Otherwise there is an infinite sequence of expressions defined by $expr_{n+1}=\neg expr_n$.