Behold A Pale Horse

There is an alternate numbering of the beast system based on only the first four seals of revelation. This is central, as the existence of the open four seals permit corporations to take part in the process named Death, upon which Hell follows soon after. The construction on this site using the groups S5 and A5 are such as to show that the Laodicean church conglomerate is spiritually deficient of life from God but the image of the beast of the ten kings emulates the carnal senses of (a) "person" with a synthetic form of logical reasoning operating upon the churches as would temptation upon the senses. The logical part of this "person" is due to the life in the image, facilitated by the ten kings but instituted by the false prophet spirit which binds the doctrine of the churches to the leavening presence of the pseudo-christian synagogue of satan rather than the bible. The five "foolish virgins" go to those that "buy and sell" doctrine in the form of this "image-corporation" construct.

Anyway, we may use the first four seals to calculate the number of the beast very efficiently. We have seen that when the fourth seal is opened,

g = 1 = (a,c,e) (a "bow" , not a subgroup)
e = (a,c,g)
c = (a,e,g)
a = (c,e,g)

Now, we wish to deliberately assume to sever our elements from the right hand ultrafilter, So we never enter "Christ" as a K4 subgroup of any octal. which we can do by counting in this fashion.

There are 35 ways to choose a triple (for our bow) from 7 symbols. Seven of these will be in our right hand (sun) octal, leaving us 28 triples in the "stars" of night. (Including our "bow"). Then there are four choices for unity to be paired with the 'bow' - as our "wine" above. That gives us 112 so far. (28 x 4 = 112) - This choice of unity also uniquely determines the static subgroup.

We note that those triples outside our "sun octal" are all different: we have generated in our four possible "left hands" all of the remaining 28 triples of the 35.

(Alternatively, note that there are in our "sun octal" seven choices for static subgroup with four possibilities for antichrist bows per static subgroup: these seven sets of four bows are disjoint, so we actually have every bow possible in opposition to our seven potentially static subgroups.)

Now we must note that for each of our 28 bows outside of our sun octal, for each bow there are four further choices for static subgroup, giving us 28*4 = 112 possible octals across all 28 bows. (We are not considering anything except that a bow may be static with one of four choices of unity).

Then one of these unique combinations, corresponds to our sun octal we started at! clearly if the correct bow is static and the correct subgroup as in the sun octal also, we have to admit that out of our 112 possible combinations of bows, one association is already valid in aligning to the sun-octal: So we may subtract 1, to get 111.

We could merely observe that of the 112 possible bows, the octal as a field itself, must align to one and only one of them; since whatever the state of the arbitrary GF(8) underneath these 112 possibilities, one of the 112 must be "correct", so we may subtract one, and consider only the remaining 111.

We now note that for every static subgroup-bow pair the seven cycles (below in pairs) are the only ones that hold one group ([b,d,f] here), and one bow, ({c,e,g} here). with one unity (a=1) here also.

(a,b,d,g,f,c,e) shares the same octal with (a,b,f,g,d,e,c)
(a,b,d,g,f,e,c) shares the same octal with (a,b,f,g,d,c,e)
(a,b,d,c,f,g,e) shares the same octal with (a,b,f,c,d,e,g)
(a,b,d,c,f,e,g) shares the same octal with (a,b,f,c,d,g,e)
(a,b,d,e,f,g,c) shares the same octal with (a,b,f,e,d,c,g)
(a,b,d,e,f,c,g) shares the same octal with (a,b,f,e,d,g,c)

Then we may extend across all possible octals by a factor of six to get 111*6 = 666.

Then we have the rock solid result that all the possible configurations of antichrist bows in our wider "sea" outside the sun octal total to 666. The beast is numbered.

In essence we are not excluding six octals, but those octals that may hold a particular subgroup static. In removing the six octals that hold fixed one static subgroup with a particular choice of unity, we either subtract 112 -1 = 111 as before, or 672 - 6 = 666 as after.

Ignore the original or "starting" octal for one moment, and restrict yourself to considering the structure of the first four seals. We are counting the number of configurations of the cosets to whatever subgroups Christ may be considered to "sit at the right hand" in.

Each K4 subgroup is in six octals, and we consider only those subgroup-triple pairs in those octals that hold that subgroup static: and we do so considering only one element as unity.

From the perspective of the static (christ) subgroup, the cosets of "other christs" are configured in 666 alternate fashions, aside from the one configuration in the original "sun octal" or field that is genuine as the "right hand". Thus "anti" takes its proper meaning "in place of". The meaning of "against" is implicit with the nature of the cosets themselves.

Thus 666 is the number of "antichrist" rather than that of GF(8). (We take Christ to correspond to one possible subgroup with one particular unity.)

To see that the six sets of 111 subgroup-triple pairs on those octals are "disjoint" we may note that there are 28 bows per octal, each under a disjoint set of seven cycles. Then 6 possible octals per subgroup-bow pair with four choices of unity each gives us a total number of 672. We then subtract the original "Christ" subgroup's six octals leaving us 666. That there are different octals underneath clouds the issue, but the fact that the seven cycles over them are disjoint sets shows they are essentially different.

Continue To Next Page

Return To Section Start

Return To Previous Page