Six Hundred Threescore And Six
Satan has set up the image system to counterfeit the floating unity of election to Jesus Christ. the 120 choices of three from ten kings may apply to any one of the remaining seven being as "facilitator" and the system of the remaining six kings that uptakes on the antichrist or "Death" spirit's 'leaven' follows with one mind as "Hell" after.
This may seem a little hard to explain, indeed it is. The one king remaining of the seven is arbitrary, because the role of facilitator is to ensure the process, and the three that are in the " balances" are thesis and antithesis (g and g^-1) and the third being 'synthesis' or (H in gHg^-1) is directed by the facilitator or fourth "bow" or pale horse as to choosing the factoring from G to H (S5 to A5).
Every element of S6 is derived to be as 30 (right) cosets of the group S4. Every octal must have its four earthy elements in some coset of S4 in S6 once we fix the static subgroup. Twelve elements of S4 are non-construcible in the octal: we must either move octal or change operation. There are properly 666 possible configurations of the pale horse "bows" in the octal for a particular octal as the sun octal.
In choosing a sun octal we require a floating unity - the choice of five refining churches is based on a fixed unity as laodicea. We need to show a bijection between choices of pale horse systems from a set of four arbitrary balances to the specific choice of the image system in place.
However we must expect to arrive at the result that the beast or 666 is actually an empty set.
S6 is reduced to (with a fixed choice of unity) 30 octals of possible left hand bows, i.e. if a=1 then there are always four possible antichrist "bows" no matter the octal or seven cycle used. These 30 octals with their bows correspond to the elements in each coset of S4 in S6. (we note the bows must not contain unity.)
There are six seven cycles holding each bow fixed over any given octal, whether in left or right. (the seven cycle or its inverse).
the sum (35-7)*4*6 - 6 = 666 calculates the number of pale horses as 666.
the sum 120*6 - 60 + 6 calculates the number of the image system as 666.
In the image we are not interested in the choice of unity but only in discovering the leavening from the source of false doctrine. The image system is concerned with equating g and g^-1 with two bows of antichrist and H as the city of "sodom or egypt" as mentioned in revelation. We require an equivocation more than a direct bijective correspondence.
There are 28 possible ways to construct triples from seven elements without choosing a subgroup of our "sun octal". There are always four possible bows (see *) for each choice of unity, we require not to choose one of four bows but one pair of six from four bows (also see *). (we must also note we require only one unity of a choice of seven!) Then we further require that there are six possible groups of seven cycles that hold any particular "bow" fixed. (We must note that there are some seven cycles holding both "bows" fixed. - One a static subgroup of the right hand and one a triple from the left. - We have excluded the right hand elements to begin with so we are ok to proceed.)
* we must accept that choosing one pair excludes another pair from the six possible pairs, (we leave behind two elements that are essentially as made by the same choice. Hence we may divide by two.) In choosing two bows from four we also separate the remaining two bows as a pair in the same octal structure (There are always four bows.) Likewise the "double minded man unstable in all his ways" is able to switch the "oil and wine" to facilitate. (This is seen in the transpositions used to factor to A5).
Lastly, we note that there are six octals containing each bow as a subgroup rather than merely a static triple: We have six octals containing one bow and six the second. Two of the first set are in the second, giving us a total of 10. We now can finish by stating that the bows of left and right are in an order that matters! (one must be g and the other g^-1). We have 120 possible pairs of bows, but their roles are such that if order matters then the inverse property of g and g^-1 permits this.
(28/7)*6/2*10= 120. (which correspond to the two pans of the balances of the ten kings as g and g^-1)
In choosing half of the possible elements (we divided by two after realising we had left a second pair behind our selection of thesis and antithesis) We now acknowledge we have a further 120 of the total 240. However one half of this is made into the number of possible choices for our synthesis. There are 60 choices to represent an H in the centre, and another 60 remaining that are paired with it! The last 60 follow uniquely as the result of the pale horse that contradicts the method as a substructure in the octal.
So there are 120 choices for "g or g^-1" and 60 possible bows that may collectively form members corresponding to the set H closed under conjugation of elements g, with a remnant of 60 determined or formed that may act as a second group or alternate A5 by the trasposition between "oil and wine". If oil were switched for the wine we would have a system wherein the dialogue had changed to show its consensus to be valid as if it were doctrine, or with doctrine as the starting consensus. Either way the dialogue is leavening.
All we need note now is the system is present within S6 with its 720 elements. We constructed every possible bow in the pale horse to total to 666. Given any generalised octal, there are 7! = 5040 possible seven cycles upon it, in fact there are 6! = 720 since the starting element in the labelling of a 7 cycle is arbitrary.
Across all our octals and seven cycles (aside from the Sun octal) we noted that there is a switch out of philadelphia with the source of leaven. We generalised all of our choices up to a choice of floating unity, so we may assume that the number of possible bows in "any octal" corresponds adequately to the 720 possible permutations in S6. In removing down from S5 to A5 we acknowledge that the synagogue is present as if it were sin in "the man", and that the seven cycle element of the "sardis" lampstand in the angel's circuit is the entry point for the leaven as from the fallen star with the key to the bottomless pit. If we count bows as if we were to also permit those in the sun octal we would have the calculation
(35/7)*4*6 = 720
Giving us our elements of S6.
The set H corresponding to sixty possible bows is closed under permutation conjugation in the image, we note that any pair of bows results in a synthesis with any other choice of bow (the third trumpet) so the division between the synthesis and the result of the unique pale horse bow (of contradiction) is such as to divide 60 elements (useful to the task) from it's remaining (contradicting) set of a further sixty. (we may choose up to thirty *pairs* before we start to drop bows as we gain them.) The result is almost a continuous or magic roundabout once the dialogue has continued that far.
We note that the division between these two sets is equivalent to the factoring down from S5 to A5 by means of the possible transpositions of the serpent spirits. We do so by say switching out the octal elements (b,c) from our familiar sun octal with those (e,f) of the earth from the left hand. (under the correspondence [a,b,c] = a = [a,e,f] in the "dead witnesses" octal pair seen before.) Likewise up to choice of left hand octal, we may transpose any pair of "earthy" elements and the serpents then counter as the third woe by simply switching oil and wine, (as unity and the synthesising wine at the table of the pale horse).
The third woe is then properly an attack of satan upon the life and resurrection found in Jesus Christ. In the resurrection (of) Jesus' K4 ultrafilter it is seen to switch out (bc) for (ef) and vice versa, yet the beast and its image switch oil and wine in the process of the dialectic - the first resurrection is under all out assault by the dragon and his two beasts.
Further the remaining two bows left behind by our choice of pairs is the spiritual "flip flop circuit" of our double minded man. The result that A5 is a normal subgroup is split into a group and coset each of sixty elements. We note that there are six possible sets (in a partition) in S6 for elements (once having factored down from S5 to A5) and then to conjugate the group A5 amongst A6 in S6 - as if it were akin to the six wings of all four seraphim before the throne of God in heaven.
Those last six elements we add to the image are the same six elements we choose to factor down by to place the synagogue of satan within the laodicean conglomerate to "work in" the source of leaven (and only the identity "e" corresponds to the Laodicean Church itself as without the source of leaven): They are the six elements of the "bows" that are formed by the dialogue process. Any two of the "current four bows" under dialogue may become thesis-antithesis and/or oil-wine. (giving six choices) The dialogue to consensus is about ruminating with the BS of the world whilst the truth is trampled under by the process itself. The choice of thesis and antithesis from the four horsemen or "bow" devices are chosen by the facilitator or the "seventh king" in place of Christ (antichrist). The process ensures that the discussion never decries the method itself, but that the same maddening wine is to be drank by all.
Continue To Next Page
Return To Section Start
Return To Previous Page |