Yes, I have meant to correct this post. I haven’t quite given the axiomatization that I had used.

]]>If memory serves (I haven’t unpacked them yet) the model and its theory are covered in …

“Model Theory” by Chang and Keisler

“A Course In Mathematical Logic” by John Bell, Moshe Machover

… so what follows is not original.

Models of T are as you described, a standard part isomorphic to (N, S, 0) and zero or more disjoint Z-chains. But letting T0 denote the three axioms given, models of T0 have a standard part isomorphic to (N, S, 0) and a possibly empty nonstandard part on which S acts bijectively — a permutation. So finite S cycles are allowed by T0, for example 1-cycles (fixed points) x = Sx, 2-cycles x and Sx distinct with x = SSx, etc. (along with the Z-chains if any).

You get a complete theory T1 by taking T0 plus the “no finite cycle” scheme ~(x = Sx), ~(x = SSx), ~(x = SSx), etc. You implicitly used this scheme when you wrote

“We may furthermore assume that x appears on only one side of each atomic clause, since otherwise the statement is either trivially true as with SSx = SSx or Sx ~= SSx, or trivially false as with Sx = SSx.”

T1 is complete because it has the same models as the complete theory T of the standard model, or alternatively because it has no finite models and is categorical in each uncountable power.

Another complete theory is T2 = T0 plus the induction scheme, as the induction axiom for (say) ~(x = SSSx) easily proves the “no finite cycle” axiom (x)~(x = SSx). To prove T2 using T1 though I assume you have to write out the inferences used in applying the elimination of quantifiers.

Finally let A be the conjunction of a hypothesized finite axiomatization of T. Then let “M sub n” be the model with standard part plus a single n-cycle, for n = 1, 2, 3, …. Then each Mn is a model of ~A, but their ultraproduct modulo a non-principal ultrafilter over the positive integers models A, contrary to the Łoś ultraproduct theorem.

—

Dan D’Eramo

The reason is that V=HOD is equivalent to the existence of a definable global well order in ZFC. If V=HOD, one can use the HOD order, and if there is a definable order, then there is such an order in type Ord, and every x is the $\alpha$th element for some $\alpha$, so V=HOD.

V=L implies V=HOD, but only V=HOD is equivalent to the existence of a definable well ordering.

]]>Thanks for this reference, Ali. I’ll update the post to mention it.

]]>Oh, I wasn’t proposing to formalize the example of the sad juggling clown, but rather to use that example to illustrate the nature of definite descriptions in a way that would help us to provide a formal account of the iota operator semantics in first-order logic.

]]>The reason the trick doesn’t work in chess is that one cannot move around the zig-zag corners on a single move (but in draughts, this is no problem). Thus, the possibility of a draw by infinite play enters the picture, which ruins the analysis. Meanwhile, in 3D chess, Cory Evans and I proved that we can put the branches on separate layers, keeping them straight, and the analysis works.

]]>I also wonder why the same trick doesn’t work in infinite chess. It seems we can embed the full binary tree in chess as well. In 3d version there was a black bishop in a layer that could unblock a finite straight path for the black king to pass. Maybe there is a way to put a number of bishops so that they can unblock a zigzag path for the king (in a single plane).

]]>I’m sure you are right. I tried to have a full discussion of the easy cases, with piles of height up to 3, since this is a case that even very young children can learn (I have taught this to classes of first-graders). But then deriving the strategy for the general case makes the whole discussion somewhat longer. Of course, it is possible to describe the strategy and proof in just a few sentences: make a move so that when you represent the pile heights as sums of distinct powers of two, then the number of powers of two appearing overall is even; if this is possible, you will win, since any move on a position with this feature will lack it, and so your opponent will never be able to reach the empty (winning) position; and if a position lacks the property, then there is always such a balancing move.

Perhaps you will prefer the discussion of Nim in my book, Proof and the Art of Mathematics, where I treat this game in the chapter on the theory of games.

]]>