By second-order reachability I mean a second-order lifting of the reachability problem on first-order structures. So let $R(X,Y)$ be a second-order binary predicate (i.e. it links a set of elements $X$ to another set of elements $Y$. Then we have some first-order predicates $P(x)$ and $Q(y)$.

Lifting how I would express reachability in second-order logic, I can express this second-order reachability in third-order logic:

$$ \forall \mathbb{P} \bigl(\mathbb{P}(P) \land \forall X Y ((\mathbb{P}(X) \land R(X,Y)) \to\mathbb{P}(Y)) \to \mathbb{P}(Q)\bigr) $$

But is this needed? Is this second-order reachability problem expressible in second-order logic, or do I need third-order quantifiers, similarly to how first-order reachability needs second order ones?


1 Answer 1


In fact, we can always stay in the second-order realm (as long as our "base structure" is infinite)!

First, let's look at the particular case where our base structure is $\mathcal{N}=(\mathbb{N};+,\times)$. Over $\mathcal{N}$ we can code finite sequences of sets as individual sets as follows: the set $X\subseteq\mathbb{N}$ codes a finite sequence of sets iff every element of $X$ has the form $2^x3^y$ for some $x,y\in\mathbb{N}$, and there is some $m$ such that whenever $2^k\vert u\in X$ we have $k<m$. The idea then is that such an $X$ codes the sequence whose $i$th term is $\{j: 2^i3^j\in X\}$.

Now suppose $R$ is a second-order binary operation on $\mathcal{N}$. Given $A,B\subseteq\mathbb{N}$, we have that $A$ is $R$-connected to $B$ iff there is some $X$ which codes a finite sequence of sets $(X_i)_{i<m}$ with the property that $X_0=A$, $X_{m-1}=B$, and for each $i<m-1$ we have $X_iRX_{i+1}$. This is purely second-order over $\mathcal{N}$, so we've avoided third-order quantification.

OK, but what if we're looking at a very low-expressive-complexity structure, such as a(n infinite) pure set? Well, we can still do the job!

  • In pure second-order logic, we can determine when a set $X$ is finite ("Every well-order on $X$ is also a co-well-order" - and if you want to avoid reliance on the axiom of choice here, you can also include the clause "$X$ is well-orderable," which interestingly is optimal in a precise sense).

  • We now can express "$A$ is $R$-connected to $B$" as "There is a pairing operation $\langle\cdot,\cdot\rangle$, a set $X$, and a binary relation $\trianglelefteq$ such that

    • the set $L:=\{x:\exists y(\langle x,y\rangle\in X)\}$ is finite,

    • $\trianglelefteq$ is a binary relation on $L$,

    • if $x$ is the $\trianglelefteq$-least (resp. greatest) element of $L$ then $\{y: \langle x,y\rangle\in X\}$ is $A$ (resp. $B$), and

    • if $x,x'\in L$ and $x'$ is the $\trianglelefteq$-successor of $x$ then we have $$\{y: \langle x,y\rangle\in X\}R\{y: \langle x',y\rangle\in X\}.$$

Note, though, that there was a cost: over $\mathcal{N}$ we just needed to quantify over sets, whereas here we need to quantify over relations. This is unavoidable: monadic second-order logic is not strong enough to perform this sort of construction without a sufficiently rich base structure.

  • $\begingroup$ Thanks, this is really interesting! How many quantifier alternations do you think are needed to express all these things? $\endgroup$ Sep 17, 2023 at 6:38

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