Question: How many fragments of second-order (classical/intuitionistic) logic are there using only the quantifiers $\exists$/$\forall$?

Specifically only those "fragments of second order (intuitionistic/classical) logic" that are also extensions of first order (intuitionistic/classical) logic.

Bonus: Answers about the "expressive reducibility" (syntactic and/or semantic) of these fragments to other logics, or vice versa, are interesting to me. However I do not understand mathematical logic (and/or model theory) well enough (yet) to phrase the corresponding questions precisely enough.

Clarification of intent 1:
Basically, everything that is well-formed in first-order (intuitionistic/classical) logic (e.g. quantification over variables), plus possibly (or possibly not) quantification (of some kind) over (some) relation and/or function symbols.

For example, the fragment discussed in this question would be excluded.

Clarification of intent 2:
The intent is to exclude fragments of second-order (intuitionistic/classical) logic formed by adding "exotic" quantifiers (e.g. like the Haertig quantifier) to first-order (intuitionistic/classical) logic.

So, for example, the fragment discussed in this other question would also be excluded.

Background: It seems like quantification can be restricted to higher-order symbols of a fixed arity and/or relation vs. function type, at least based on the definition of monadic second order logic (discussed e.g. here or here) which seemingly only allows quantification over unary/$1$-ary predicates, but not over predicates of any other arity, nor over function symbols of any arity.

The Plato encyclopedia discusses fragments of second order logic that are extensions of first-order logic, see here. However it seems to discuss fragments formed using "exotic" quantifiers like the Henkin quantifier or the Haertig quantifier.

The Plato encyclopedia also mentions that Shelah has classified "nice" fragments of this kind (at least using "exotic" quantifiers) into four classes up to bi-interpretability, a result also mentioned in this answer on Math.SE.

(Presumably there are fewer equivalence classes using the "weaker" notion of mutual interpretability? With mutual interpretability being "weaker" because ZF and ZFC are mutually interpretable but not bi-interpretable?)


1 Answer 1


This is community wiki because it is my own attempt to answer the question and it is probably wrong.

Classical logic case: The previous attempt failed to consider how every $(N-1)$-ary function can be interpreted using a $N$-ary predicate, with such interpretations corresponding to "special" $N$-ary predicates.

In other words, quantification over $N$-ary predicates should probably (is this actually true?) imply quantification over $(N-1)$-ary functions, but not vice versa.

So considering ($0$-$N$)-ary predicates and functions, we have the special cases of $0$-ary predicates and $N$-ary functions, each of which will give $2$ options (quantification allowed or not allowed).

But then for $1$-ary predicates through to $N$-ary predicates, and $0$-ary functions through to $(N-1)$-ary functions, based on the above comment it would seem that we have three options for each $n=1, \dots, N$:

  • allow quantification for $n$-ary predicates (and thus also for $(n-1)$-ary functions)
  • allow quantification for $(n-1)$-ary functions but not for $n$-ary predicates
  • allow quantification for neither

Thus the upper bound bound for the total number of distinct fragments would actually be $2 \cdot 3^{N-1} \cdot 2$.

Using the notation from the previous attempt, this would then correspond to an upper bound of $2 \cdot 3^{ \min\{r - 1 , f-1 \} } \cdot 2^{f - (r-1)} = 2 \cdot 3^{r-1} \cdot 2^{f-r+1} $ in the case that $f \ge r$,

and when $r \ge f$, well then the question is not necessarily well-defined because again it probably is not possible to permit quantification over $N$-ary predicates without also permitting quantification over $(N-1)$-ary functions.

Intuitionistic logic case: Presumably the same idea would still apply as before, namely that one would need to multiply the upper bound for the classical logic case by an appropriate power of $2$ (because allowing / disallowing universal quantification becomes a separate/distinct question from allowing / disallowing existential quantification).

Previous Attempt:
Classical logic case: In classical logic, for each kind of higher-order symbol we have $2$ choices: allow (existential/universal) quantification, or don't.

So that should give us $$2^{r+1} \cdot 2^{f+1} $$ fragments of 2nd order classical logic that do not allow any quantification over relation symbols with arity greater than $r$, nor allow any quantification over function symbols with arity greater than $f$.

($r+1$ and $f+1$ instead of $r$ and $f$ by including $0$-arity symbols.)

Technically $2^{r+1} \cdot 2^{f+1} - 1$ if we are to exclude FOL itself.

Intuitionistic logic case: In intuitionistic logic, the quantifiers $\exists$ and $\forall$ are "not equivalent", i.e. cannot be defined in terms of one another (cf. e.g. the [Plato Encyclopedia][3]), so we have $4 = 2^2$ choices for each kind of higher-order symbol: allow no quantification of any kind, allow existential $\exists$ but not universal $\forall$ quantification, allow universal $\forall$ but not existential $\exists$ quantification, or allow both kinds of quantification.

So then it seems that there should be $$4^{r+1} \cdot 4^{f+1} = (2^2)^{r+1} \cdot (2^2)^{f+1} = 2^{2r+2} \cdot 2^{2f + 2} $$ fragments of 2nd order intuitionistic logic that do not allow any quantification over relation symbols with arity greater than $r$, nor allow any quantification over function symbols with arity greater than $f$.

Again technically $4^{r+1} \cdot 4^{f+1} - 1$ if we are to exclude FOL itself.


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