Instead of steam sterilization I would try to bury it under 50cm of mulch. Ripping out the rizomes requires care and only makes sense in very loose soil, otherwise you just multiply it. ]]>

How do I know it’s correct? Because such a famous and shocking theorem has been examined by every expert in the field, and any one of them would be happy to become famous by spotting a flaw in it.

]]>**Definition.** A choice for player A **dominates** a choice if

for all

**Definition**. We say a choice is **dominant** if it dominates *all* other choices for that player. In particular, choice for player A is **dominant** if

for all and all

]]>is the security value for player B when that player is using a maximin strategy! (The minus sign here shows up because we’re dealing with a zero-sum game.)

I don’t know what an “intersection strategies” is. By the time we get to Theorem 5, we’ll see that a Nash equilibrium *always* exists for the kind of game we’re talking about. But maybe this will answer your question: the equation

says that the security level for player B when they are playing a maximin strategy is minus the security level for player A when *they* are playing a maximin strategy!

I mean, I’m not in the least bit agitated; I’m a dyed-in-the-wool humanist and a radical constructivist whenever it doesn’t make my life unnecessarily difficult (I use the Axiom of Infinity to prove existence in the paper I’m working on, sorry Kevin Knuth), so I find the whole situation ecstatically hilarious. Although Alain Connes could very well put a contract out on me; I’ll have to stay away from NYC, he has friends at Columbia!

]]>“From this, the model itself is considered recursive or not

depending on if it has recursive functions.”

And follow that with the statement just below Tennenbaum’s Theorem on page 11:

“Thus, an explicit line is drawn between the standard model of PA and all of the countable non-standard models of PA, as any attempts to formulate arithmetical operations in such a non-standard model will be severely restricted.”

This is NOT true as I can very well demonstrate! Let N* be the set of all hyper-naturals, then any hyper-natural n* takes the form, m.p, where m designates the von Neumann ordinal, which I refer to as the number of “base elements,” and p designates the number of hyper-elements, i.e. the number of times the von Neumann ordinal has been reflected, so to speak. Let m.p and n.q be arbitrary elements of N*, then:

m.p + n.q = (m + n).(p + q); and

m.p * n.q = m.p * n. m.p * q

= m * n.p * n. m * q. p * q

= m * n.(p * n) + (m * q) + (p * q)

And I can use the standard parametric version of the Recursion Theorem to prove existence of these functions in a very straight forward way. Whether or not I need to use transfinite recursion is really irrelevant to the issue at hand! I mean, recursion is recursion!

I’m writing up a paper and what I prove in that paper is that there exists a counter-example to Tennenbaum’s Theorem. It’s not that I need to add an additional constant, or anything else for that matter, to the signature, it’s just that there does not exist an isomorphism between (N, <) and ω^2, as I understand things, because every well-ordering is isomorphic to a UNIQUE ordinal!

I’m writing up a paper and there are those who will accept the result! And the thing is, even with my ignorance, I can see how useful these numbers can be! You can extend the hyper-naturals to hyper-integers, the hyper-integers to hyper-rationals, using Cauchy Sequences, the hyper-rationals to a novel set of hyper-reals, and the hyper-reals to the hyper-complex. Think about it . . .

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