Desperately Seeking Integers
A few twists on Turing’s proof of undecidability of predicate calculus
By permission of Rich Longmore, artist, blog source 
Alan Turing presaged Stephen Cook’s proof of completeness of Turing reduced the halting problem for his machines to the decision problem of the firstorder predicate calculus, thus showing (alongside Alonzo Church) that the latter is undecidable. Cook imitated Turing’s reduction at the level of propositional calculus and with a nondeterministic machine.
Today we present a version of Turing’s proof and show how it answers a side issue raised in our previous post.
Completely aside from the ongoing work by Harvey Friedman previewed in that post, we—that is Dick and Ken—were interested in the boundary between decidable and undecidable theories in logic. We noted Michael Rabin’s theorem that the theory of the rationals with order is decidable with firstorder quantifiers and secondorder quantifiers allowed only over unary predicates—that is, over subsets of We sketched how quantifiers over ternary predicates—that is, over subsets of —enable defining properties of and from which undecidability follows by the mentioned theorem of Julia Robinson.
This left the binary case. Most in particular, we were interested in the decision problem for secondorder sentences of the form
where the only predicate symbols in (besides and ) are and has only firstorder quantifiers. We were given references to papers by László Kalmár and William Quine but judging from this by Church and Quine (see page 183 and also this and this plus these slides) we wonder what extra is involved, for instance pairing. Our search for a crisp and brief proof led back to Turing’s firstorder theorem.
Defining a Stairway
Our starting point is to build a predicate so that will define a set with properties needed of the integers. Our intent is to make define a kind of ceiling function the least “integer” The first rules we write say that is a functional relationship:

For each there is a unique so that :
and
Thus defines the desired function The further rules we need are:
 The function is nondecreasing.

The function is a “steplike” function:
This means that is constant on an interval.
 We add one more rule for good measure but it does not do all we might expect:
 The function tends to infinity:
Now all we need is to define the successor relation via more rules and postulate an element to act as zero. The rules for successor are:
Among several ways to handle we find it most thematic is to leave it unquantified and assert This will be on the lefthand side of implications of the form Any interpretation—that is, any assignment of a function on for and a value for —that makes all the parts true will make the bottom of a stairway that has infinitely many steps above Let be the conjunction of and all the above rules.
The last rule for might seem to imply the stairway climbs to infinity. As we’ll discuss, this is not enough to make a single “stairway to heaven.” But it does provide that any angel who climbs some number of steps above will need exactly steps to get back down to
Encoding Computations
Proofs of Turing’s theorem such as this implicitly use relations like and for binary strings on Turing machine tapes, which numerically work out to and We will use a universal twocounter machine instead. We don’t need the arithmetic details of why it is universal, such as Marvin Minsky’s coding trick. The machine input is in unary but we can just define it as a natural number To represent the current state and the counters at any time , we use three more binary predicates:
 the machine is in state at time
 counter 1 holds value at time
 counter 2 holds value at time
We can write rules to make these functions of The program has a finite number of states and so we can (if we wish) use dedicated symbols for them. We can take as the start state. Following Minsky, we need only two kinds of instructions:
 : If in state , increment counter and go to state
 : If in state and counter is zero, go to state , else decrement counter and go to state
Each state has exactly one of these instructions except the halting state The rules corresponding to these instructions all have the form
where is
In case the instruction for state is we have as
whereas in case the instruction is a conditional jump we get as
We conjoin all the machine rules into one finite formula The conditions to start in state on a given input value are:
Here we mean to be a fixed natural number but we can’t say so literally, so what includes are terms until we get to the fixed value Finally, the condition to halt eventually in state is:
Turing’s Theorem
Now given any number , take
We want to decide whether is valid. Now is firstorder: we haven’t put in the existential quantifiers over the predicates yet. To be valid in our setting means that whatever relations on are used for the four predicates and whatever value is used for , the resulting statement about rational numbers is true.
Theorem 1 The formula is valid if and only if on input halts in state Hence firstorder predicate calculus is undecidable.
Proof: First suppose halts. We need only consider interpretations that make true. Such interpretations set up the initial configuration of with standing for zero and make every instance of the universally quantified implications implications in true. The chain of those instances eventually yields where is the number of steps takes to halt. Thus and hence overall are made true.
Conversely, suppose is valid. Then we can pick the interpretation that makes the actual integer ceiling function and This makes true so it must make true with ranging over the positive integers, which means that must halt.
We seem to have sacrificed some generality by basing on , but in fact we used nothing about the rationals except being a total order (including trichotomy: ). We could axiomatize those properties too. Alternatively, we could forgo equality in favor of having function symbols as in the Turing machinebased proof mentioned above.
Ups and Downs in the SecondOrder Case
Now we put the secondorder quantifiers on the predicates in order to prove our desired theorem:
Theorem 2 Consider sentences of the form
where is firstorder. The problem of deciding whether is true in is undecidable.
A natural impulse is to take for a given input as before. This leads to a notable issue, one that Ronald Fagin reminded us of. If halts, we get that (for that ) is true. Indeed—and this is a first hint that wires are crossed from the above proof—we can use the genuine integer ceiling function for the “” part and the rest follows.
Conversely, suppose does not halt. The rub is that we can still make true. Take such that the derived set , our set of “integers,”includes Take Then is the successor of for all but is not the successor of anything. Although the values of for are fixed for all by actions of the machine, nothing prevents us from blithely defining to be true. This makes true too.
We forgot about the rule that grows unboundedly but it doesn’t help: we can just add the rest of the natural numbers to Attempts to fix this by running backwards from state are forestalled by the possibility of adding to for all as well—our axioms are too weak to prevent having infinitely many predecessors.
The upshot is that although we can will into existence multiple infinite staircases, we cannot guarantee a single staircase that goes to heaven. In broad terms, we cannot rule out taking a nonstandard integer number of steps to halt (for connections to start here). But in this situation we can fix the problem by a simple twist (used also here).
Proof: Define by asserting instead that doesn’t halt. That is, take
If doesn’t halt, then the standard assignment to giving and works for that as well. The converse is now to suppose that does halt in state We need to show that there is no way to instantiate the four binary predicates and to make the resulting sentence true. By it sets up the initial configuration. The part leans only on features of that the instantiation must make true. Thus successive configurations are forced by the implications in to follow correctly, so that is made true for some that does belong to the staircase above This immediately contradicts of So is false.
Thus is true if and only if does not halt.
This finally answers the last post’s desire for showing concretely that secondorder existential is undecidable with binary predicates, needing no reference other than Turing (OK: and Minsky). We would still like to trace this historically through the abovementioned references to Kalmár and Quine and perhaps others. In contrast to great past attention to minimizing the alternation and number of quantifiers, we are curious also about minimizing the number of binary predicates. We note how this 1984 paper by Warren Goldfarb gets the number of binary predicates down to one but at a greater cost in unary predicates, of which above we’ve used none. We are also not sure how our setting with domain and a fixed meaning for plays in the tradeoffs.
Open Problems
What is the best reference? Is there a nice and clean proof for the case of just one dyadic predicate? What other tradeoffs, such as between function symbols and predicates using equality (or existsunique quantifiers), are in play here?
[added reference to Goldfarb paper and changed next sentence; changed picture at top with permission notice]
Trackbacks