Googology Wiki
Register
Advertisement
Googology Wiki

(Note that a regular user in wikia is not able to keep harassers and web trolls away, so please ignore if some web troll who thinks nearly everything is "meaningless" replies this post)

Define a sequence of first order language Rn as: Rn has n+1 2-ary relation symbols {<, <1, ..., <n}. Any Rn model can be regarded as an Rn+1 model. Define Rω has 2-ary symbols {<, <1, <2, ...}

For any BMS matrix M with n rows, we define a finite Rn-model f(M) as:

If M has m columns, then f(M) has m elements related to those columns. If a,b∈f(M) such that their corresponding columns a',b' s.t. a' is at the left of b', then set a<b. If the element at k-th row in a' is an ancestor of the element at k-th row in b' (note that ancestor is defined in the definition of BMS), then set a<kb.

Now we define an Rω model R: the underlying set is limit ordinals in ω1+1, < is just regular order, α<nβ iff Jα<nJβ, i.e. Jα is a Σn elementary submodel of Jβ. Let Jλ be the mostowski collapse of the skolem hull of the empty set in Jω₁.

R can be regarded as an Rn-model. Assume P is a finite Rn-submodel of R, and k is a positive integer such that there exist q,p such that p is the <-largest element of P, q<kp, and q is the <-largest such element.

Then we define a finite Rn-submodel of R, gk(P).

Let X=q∩P, Y=P\{q∪{p}}. Consider the formula φ(X): there exists an x, such that x=<y> is a finite increasing sequence of ordinals, and X∪y is Rn-isomorphic to X∪Y, and if k>1, then y∪{Ord} is Rk-1-isomorphic to Y∪{Ord} (in Jα, Ord=α). By VI.3.4 in Devlin's book "constructibility", φ is Σk. By Jp⊨φ(X) we have Jq⊨φ(X), let x=<y> be its J-least witness.

Define g(P) be the Rn-model P∪y.

Now for a finite Rn-model P such that < is a total order, define h(P) be the least ordinal α such that there exists an injective Rn-homomorphism P→R maps the largest element of P to α. If h(P) and g(P) both exist, then h(g(P))=h(P). Deleting the last element of P (note as i(P)) would make h(P) become strictly smaller.

For the n-row matrix M=(0,0,...,0)(1,1,...,1) (such kind of M is a main sequence of BMS), {λ, ω1} is isomorphic to f(M), so h(f(M)) exists.

For any n-row BMS matrix M such that M has a bad root, if h(f(M)) exists, we regard f(M) as the Rn-submodel of R correspond to h(f(M)).

Then it's easy to verify that h(f(M[n']))≤h(i(gkn'(f(M))))<h(f(M)) (so in particular h(f(M[n'])) exists), where k is the largest row index in the last column that the element has an ancestor.


Also, if M don't have a bad root, let M' comes from M by deleting the last column, then h(f(M'))≤h(i(f(M)))<h(f(M)).

Because h(P) is an ordinal, the image of h do not have an infinite descending chain, so we have proven that BMS terminates.

Advertisement