The full solution to the dog puzzle

A couple of days ago I posted A logic puzzle about dogs. Read that post first and try solving it before reading on!

Below is the full explanation of the puzzle. Heavy spoilers, obviously.

The dog society is structured identically to the Robinson axioms for natural number arithmetic. Dogs are numbers, Spot is 0, the alpha of n is n + 1, the referee for n and m is n + m, the counselor for n and m is n × m, and the strength relation is the < relation. This means that “the marriage counselor for Spot’s alpha and the referee of a fight between Spot’s alpha and Spot’s alpha” is translated to 1 × (1 + 1) = 2, which is Spot’s alpha’s alpha. In Robinson arithmetic, you can also prove that ∀n (n < n + 1).

As for the question of if it’s possible for a dog to be stronger than Spot, Spot’s alpha, Spot’s alpha’s alpha, and so on: The primary difference between Robinson arithmetic and Peano arithmetic (the usual axiomatization of natural numbers) is that the Robinson axioms have no induction axiom (which would be something like “If Spot has a property, and if the alpha of any dog with the property also has the property, then all dogs have the property”). The induction axiom serves to rule out many models of the axioms that are not actually the natural numbers.

If the induction axiom is stated using second-order logic, then the axiom system uniquely pins down (ℕ,+,×,>) and there are no dogs besides those in Spot’s hierarchy. But the induction axiom cannot be stated as a single axiom in first order logic, since it involves quantifying over all properties. For first-order Peano arithmetic, we instead have an infinite axiom schema, one for each property that is definable within the first-order language. This turns out to be strictly weaker than the single second-order axiom, as there are some properties of numbers that cannot be described in a first-order language (like being larger than a finite number of numbers).

What this amounts to is that first-order Peano arithmetic with its infinite axiom schema is too weak to pin down the natural numbers as a unique model. There are what’s called nonstandard models of first order PA, which contains the ordinary numbers but also an infinity of weird extra numbers.(In fact, there exist models of first-order PA with every infinite cardinality!) Some of these numbers have the property that they are bigger than all the regular natural numbers.And since Robinson arithmetic is strictly weaker than first-order PA (lacking an induction axiom as it does), this means that Robinson arithmetic is also not strong enough to rule out numbers greater than all elements of ℕ. Which means that we cannot prove that there are no dogs stronger than every dog in Spot’s hierarchy!

I made this puzzle to illustrate three things: First, how the same axioms, and even the same models of the same axioms, can have wildly different interpretations, and that a shift in how you think about these axioms can make seemingly impossible tasks trivial. Second, how axiom systems for structures like ℕ can (and inevitably do) fail to capture important and intuitively obvious features of the structure. And third, how logic is so interesting! Just trying to create a simple rule system to describe one of the most natural and ordinary mathematical structures that we ALL use ALL THE TIME for reasoning about the world, turns out to be so nontrivial, and in fact impossible to do perfectly!

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