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Polyps:
Strongs:

A Bag Full Of Holes

With the disrupted octal from the serpent spirits of the left and right hands as seen in the case of the two witnesses lying dead in the streets of the great city we have the correspondences of the serpents of the sixth seal as follows (these are illegal in the octal, though preserve the "hail and fire mingled with blood").

a = 1 = [a,b,c]
e = [a,d,e]
f = [a,f,g]

paired with (preserving hail and fire mingled with blood)

a = 1 = [a,e,f]
b = [a,c,d]
c = [a,b,g]

Instead of the correct K4 ultrafilter which has the associations,

a = 1= [a,b,c]
b = [a,d,e]
c = [a,f,g]

We note that the seven "Euphrates octal's" under C7 from the original or "sun" octal described in the sixth trumpet may be formed by the transpositions upon the (Sun) octal (b,c) (d,e), (f,g). (but not the products in pairs (b,c)(d,e), (b,c)(f,g) and (d,e)(f,g) - which are not constructible from the valid frobenius maps on the GF(4) ultrafilter.) . This gives us only three (and not six) elements that we may transform with from the valid sun octal of the witnesses to the wider euphrates octal and back again.

We note that in the correspondences above, the permutation (b,c) in the octal are actually the permutation of elements in the K4 filter from [d,e,f,g] to [f,g,d,e]. Thus we have immediately three permutations of S4 on these elements, induced by the valid K4 filter.

We also have the case in the result of switching e<=>c and f<=>b (in the K4 ultrafilter or also the Euphrates "serpents") is also a map starting from [d,e,f,g] but mapping to [d,c,b,g]. As a map between left and right hand variants of the K4 ultrafilter this also maps the serpent sets to each other, (though switching pairs)

Ie.

a = 1 = [a,b,c]
f = [a,d,e]
e = [a,f,g]

becomes

a = 1= [a,f,e]
b = [a,d,c]
c = [a,b,g] etc.

and upon the right handed K4 filter

a = 1= [a,b,c]
b = [a,d,e]
c = [a,f,g]

becomes in the left hand, a valid K4 filter also.

a = 1= [a,f,e]
f = [a,d,c]
e = [a,b,g] etc.

So properly these switches may be used as an "automorphism" that maps one hand onto the other. By pre and post multiplication (conjugation) by this map (c,e)(b,f) we may obtain a new set of three permutations. i.e. [d,e,f,g] becomes [d,c,b,g] (premultiply) then say after (b,c) (as mapped) we map to [d,b,c,g] then lastly postmultiply to [d,f,e,g]. This is a new permutation not obtained before. However there is some sleight of hand here,.. we are indeed using (b,c) on the "wrong hand". (The serpents could only use it on the K4 form containing [a,b,c])

There is the issue here in that we are mapping

a = 1= [a,b,c]
b = [a,d,e]
c = [a,f,g]

using (c,e)(b,f) (b,c) (b,f)(c,e) to

a = 1= [a,b,c]
b = [a,d,f]
c = [a,e,g]

Which is seemingly valid yet is from a different octal to both left and right from the valid "sun" octal. (But the other resultant hand generated with a = 1 = [a,e,f] is preserved.) The action is as if the elements of the sun octal's left hand {e,f} are transposed. The truth of the matter is that the octal is not acted upon as with a simple bijection between left and right in this manner. The situation is resolved thus.

a = 1= [a,b,c]
b = [a,d,e]
c = [a,f,g]

under (c,e)(b,f) is mapped to
a = 1= [a,f,e]
f = [a,d,c]
e = [a,b,g]

And (b,c) is not a valid frobenius map on this set! However if we use frobenius to induce (e,f), ie
a = 1= [a,e,f]
e = [a,d,c]
f = [a,b,g]

And map back with (c,e)(b,f) again to,.
a = 1= [a,c,b]
c = [a,d,e]
b = [a,f,g]

So in fact, the frobenius map on one hand also induces the frobenius map in the other hand. We may not simply induce [d,e,f,g] to [d,f,e,g] as we would wish.

Note the frobenius map that holds static a = 1 = (a,e,f) in the left handed euphrates octal also cycles the three subgroups in K4 (in both right and left hands) as if C3. This is the same action of the "sun" octal's frobenius on the corresponding GF(4) ultrafilter in the octal's right hand. (this K4 ultrafilter is from any "plausible" right hand to pair with the euphrates octal's substitute left.)

Now the situation during the earthquake at the two witnesses resurrection is as:

[a,b,c], [a,d,e] , [a,f,g], with [b,e,f], [b,d,g], [c,d,f], [c,e,g] with unity (a = 1)
[b,d,f], [a,b,c] , [b,e,g], with [a,d,g], [c,d,e], [c,f,g], [a,e,f] with unity (b = 1)
[c,d,g], [b,d,f] , [a,d,e], with [b,c,e], [a,c,f], [e,f,g], [a,b,g] with unity (d = 1)
[c,e,f], [c,d,g] , [a,b,c], with [a,d,f], [b,f,g], [a,e,g], [b,d,e] with unity (c = 1)
[a,f,g], [c,e,f] , [b,d,f], with [b,c,g], [d,e,g], [a,b,e], [a,c,d] with unity (f = 1)
[b,e,g], [a,f,g] , [c,d,g], with [d,e,f], [a,c,e], [a,b,d], [b,c,f] with unity (g = 1)
[a,d,e], [b,e,g] , [c,e,f], with [a,c,g], [a,b,f], [b,c,d], [d,f,g] with unity (e = 1) On the right hand

Now note the row on the right hand list with c=1. This octal contains those three groups [a,b,c],[a,d,f] and [a,e,g]. Likewise notice the second row contains [a,b,c] also.

With (a = 1) and [a,b,c] static the three rows containing [a,b,c] correspond to the first, second and fourth rows of the seven cycle acting downwards. (thus the elements sent to themselves under frobenius in the wider set about the original octal.)

The remaining possibility is formed in the second row.

a = 1= [a,b,c]
b = [a,e,f]
c = [a,d,g]

So if a = 1 and [a,b,c] is static we reduce to the three rows;
[a,b,c], [a,d,e] , [a,f,g], with [b,e,f], [b,d,g], [c,d,f], [c,e,g] with unity (a = 1)
[b,d,f], [a,b,c] , [b,e,g], with [a,d,g], [c,d,e], [c,f,g], [a,e,f] with unity (b = 1)
[c,e,f], [c,d,g] , [a,b,c], with [a,d,f], [b,f,g], [a,e,g], [b,d,e] with unity (c = 1) i.e. (a,b,c) is static.

which are essentially the same three octals yet with [d,e,f,g] permuted.

[d,e,f,g] which permits the permutations of the K4 filter of [f,g,d,e], [e,d,f,g],[d,e,g,f]
[d,g,e,f] which permits the permutations of the K4 filter of [e,f,d,g], [g,d,e,f],[d,g,f,e]
[d,f,e,g] which permits the permutations of the K4 filter of [e,g,d,f], [f,d,e,g],[d,f,e,g].

So there are twelve permutations of [d,e,f,g] apparent that correspond directly to 12 similar permutations of the left hand.

But the composite permutation using [d,e,f,g] to [f,g,e,d] is actually illegal! In permuting [d,e,f,g] to [f,g,d,e] and then to [f,g,e,d] as we would using frobenius on the K4 filter;

a = 1 = [a,b,c]
b = [a,d,e]
c = [a,f,g]

to
a = 1 = [a,b,c]
c = [a,d,e]
b = [a,f,g]

and then cycling the filter in the octal using frobenius on GF(8) to switch to
a = 1 = [a,d,e]
e = [a,f,g]
d = [a,b,c]

In order to employ frobenius again as (d,e) to
a = 1 = [a,d,e]
d = [a,f,g]
e = [a,b,c]

And cycling again, we see that the effect is null.
a = 1 = [a,b,c]
b = [a,d,e]
c = [a,f,g]

So frobenius from the octal sustains the action of frobenius on the K4 ultrafilter. (ie employing (b,c) as frobenius, then cycling in the octal and applying it as (d,e) again and cycling back is as if (b,c) were applied twice and thus cancels out the effect.)

However this does not invalidate the action of the four left handed columns introduced by the serpents. They may indeed switch (d,e) even though [a,b,c] is static, this is the given "key" used for 'T' in C(x,Ty).


Thus we have twelve, yes, twelve permutations of S4 constructed on the four "earthy" elements {d,e,f,g} from symmetries of the original K4 ultrafilter (a=1=[a,b,c], b=[a,d,e], c=[a,f,g]) held firm by frobenius by C7 in the wider euphrates octal. (The frobenius map reveals those rows of the euphrates octal under C7 that keep [a,b,c] static, as if preserving the K4 ultrafilter.

It follows then that as satan has added only one fallen star to the earthy elements (the one with the key to the bottomless pit) to construct the city of S5 and also the human source of leaven (liars in the church, the synagogue of satan) as an element to that to construct S6 with the image, then despite these efforts there are not four intial stars to complete the "missing" twelve elements of S4; (which is the subgroup acting solely on those four earthy elements or the first four churches of the letters in revelation.) So in fact, 12 of the full 120 elements of S5 are missing, 1/10 of the city of laodicean Babylon falls in the earthquake.

The result is that these permutations are not truly induced from the trinity or K4 ultrafilter on the original octal and are not native to these four elements. The remaining elements of S4 that are indeed missing provide a hole - it (the city) is an incomplete system. This incompleteness binds satan in the bottomless pit - he may not complete his design. (Despite saying "I will..." He cannot complete his oath in his own manner even.)

In the judgement of the city of Mystery Babylon God states:

Rev 18:6 Reward her even as she rewarded you, and double unto her double according to her works: in the cup which she hath filled fill to her double. (KJV)

In effect God judges Babylon, (she whom is drunk with the blood of the saints) with giving her blood to drink. He replaces the 12 induced elements laid claim to; (falsely) by the "serpents" of the sixth trumpet with 24 native elements of S4 that close her system so that no one will ever be able to exit her, or to see God's approval. They are cut off from Him forever.

Satan likewise is bound in this system, now leavened to the state as the (shown) completely false prophet or totally leavened by false doctrine - and become the woman, the image. It burns along with the beast it rides - and God casts her into the lake of fire.

From the ultrafilter above we merely need to state that this replacement of the elements occurs when the K4 ultrafilter is removed from the former creation at the second coming and great white throne judgement - the 12 elements will no longer be merely "induced" but concrete. Then, as the elements in the former creation melt with fervent heat and the former things pass away, the saved with Christ are entered into a new heaven and new earth formed about the (to be no longer separated from everything made new about it) truly valid K4 ultrafilter.

The former creation, then truly becomes the "Death" and "Hell" together - these are the second death. And they are cast alive into the lake of fire.


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