Revelation

The Sixth Trumpet
First we note that the golden altar (associated with fire) before God has four horns. We can correspond these to the four beasts by their complements which are necessary in keeping the ultrafilter in correct correspondence. I.e. if unity is 'a' and [0,b,d,f] is held fixed by frobenius then in the octal ultrafilter of Christ, (not in the K4 of ultrafilter of seals 57)
e = [0,a,b,c]
g = [0,a,d,e]
c = [0,a,f,g]
and the second beast like wise has its fixed group [0,b,e,g] which has another horn for unity as 'a' by [0,c,d,f], also [0,c,d,g] has [0,b,e,f] and [0,c,e,f] has [0,b,d,g]
since we can permute any symbol to unity, these "horns" are not fixed.
However, if we choose a single element as unity 'a' say, then these four horns correspond to those four choices of unity and together with the complements of the static right handed subgroups (with unity together) not in the GF(4) ultrafilter, but in GF(8). Thus we can equate the four angels loosed (as also the four horsemen) as corresponding to associations with the four antichrist "bows" in the octal, rather than solely GF(4) for which they are also to a degree valid, (As hail and fire mingled with blood.) These arrangements of "horns vs. angels" are paired together as are the generators of C7 over each octal  they are arranged in four pairs for the four possible "fire" columns from each separate choice of unity.
Note that sums of subgroups within this set of four horns/angels {[0,c,e,g],[0,c,d,f],[0,b,e,f],[0,b,d,g]} using our subgroup addition as before always give a member of the following set: {[0,a,b,c],[0,a,d,e],[0,a,f,g]}. (The additive elements of the K4 ultrafilter)
Now, no seven cycle which acts on our original octal could possibly permute these two sets of subgroups in their union. However, it is a valid octal group. (it simply has the elements 'f' and 'g' swapped)
This wider set of octals which do not correspond to our eight C7 groups is a "wider way" as called "the great river Euphrates" which runs through the metaphorical "Babylon" or Chaldea  these four angels were bound in the simple unsealed system without "fire" but are now loosed and have greatly increased the number of required seven cycles to fill this wider system. (though swapping the elements f and g has only doubled the apparent required number of seven cycles.)
These four angels are "tooled up" with the locusts of trumpet five to destroy every correct seven cycle in the mystery of the golden lampstands. They are prepared for an hour (one particular labelling of the ultrafilter) a day, (The twenty four associations of left/right under frobenius) a month (the correct number of octals in the "sea") and a year (a cycle of months  a unit of seven for multiplication, changing the apparent unity.)
so (1 labelling) * (24 seven cycles/labelling) * (15 octals * 2 octals/seven cycle) * (7 choices of unity/octal) = 5040 octals
Which is equal to the order of the set of all permutations of
seven symbols  They aim to break every conceivable work of the Holy Spirit in redemption. (The third part of men  the operation of the octal or the Holy Spirit in the ultrafilter upon the singletons as in the mystery of the seven candlesticks (churches containing men).)
The number of the army John heard was 2x10^8
which is (4 associations with any choice of unity) * 5 elements (five months) of one cycle * (2 octals/seven cycle * 5 seven cycles/octal)^(7 churches or dimensions) which is 200,000,000  see the subsequent page on the mystery of "the seven angels and seven candlesticks" page for help with this number. Thus there are that many combinations of five elements in the "sea" over the original octal depending on which lampstand is before the throne of God. (Corresponding to Laodicea  Christ's victory, not the angel's)
The factor of 10 raised to the power seven is "mysterious" but remember that two singleton elements, as Laodicea and Philadelphia are present in every row and column, not just the top row with the leftmost two columns.
So, John moves on to how he saw the horses of trumpet five arrayed in this vision.
The breastplates (protection of a different but valid octal) is comprised of "fire" the groups from different octals corresponding to the horns of the altar: {[0,c,e,g],[0,c,d,f],[0,b,e,f],[0,b,d,g]}, jacinth, which is the set of subgroups also common to the original octal {[0,a,b,c],[0,a,d,e],[0,a,f,g]} (Jacinth is deep blue and corresponds to the wider river Euphrates) and he also saw brimstone.
Brimstone could only mean the fiat use of the octal itself. It also corresponds to the reward for blasphemy of the Holy Spirit. There is no seven cycle from our set of eight C7 groups that permutes these elements, (They permute the original octal) But if we use them anyway, we will find a method of "corrupting" the ultrafilter. Likewise, if we choose brimstone to permute these elements themselves in a particular way, then we have a collapse of the K4 ultrafilter.
Using a seven cycle that permutes these subgroups through an octal group, thus: (a,b,d,c,g,f,e) We can choose brimstone to be an incorrect choice of a seven cycle element, so when taking 'a' as unity the elements correspond under frobenius in GF(4) as follows.
a = [0,a,b,c] = 1
b = [0,a,d,e] = x
c = [0,a,f,g] = x^2
So the subgroup/element {a,b,c} correspond to "fire" "smoke" and "brimstone", or, fire + brimstone = smoke (which obscures the air or seven cycle)
We note that
b and c would have to be permuted to each other (switched) if a is corresponding to unity here. We have seen this before, but we will approach John's explanation and find out why we have this corruption of the ultrafilter (which can not hold in the octal  frobenius is cyclic of order three, so the transposition (b,c) is implausible)
The heads of the horses were as lions; (heads in parentheses)
a(bc)(de)(fg)
so we read from left to right  He saw the elements arrayed in the same order as the Lion of the tribe of Judah open the last three seals  corresponding to the subgroups [0,a,b,c],[0,a,d,e],[0,a,f,g] The heads of the horses are as lions. We will associate the head with the subgroup "a(bc)"
Out of their mouths issued fire and smoke and brimstone  as above we will call the element "a" the mouth, and in combination with (bc) or (de) or (fg) the fire, smoke and brimstone respectively. we will assume the operation holds on these elements akin to subgroups, so we have the association that fire + brimstone = smoke.
The third part of men (the correct binary products of GF(8)'s operations) is destroyed by this association, a contradiction following from the frobenius map.
The power is in their mouth. ('a' is not an element in the octal that can be associated with unity and belong to a subgroup held fixed by frobenius. it is simply mathematically impossible, any permutation of the elements that does so would render the set null.)
and their power is in their tails, the set;
{d,e,f,g} which is not a "head" ("a(bc)" was the head remember), but their tails are like serpents (the whole body except for the head, or mouth  the tail could be {b,c,d,e,f,g} so the whole body is all tail, except for the mouth "a".
The tails have heads, commonly [0,c,d,f],[0,c,e,g],[0,b,e,f],[0,b,d,g]  with which they do hurt. By permuting the elements other than those in the mouth, they can exchange the operation for another, using the method of the locusts to permute the elements in groups of four or five.
If [0,a,b,c] corresponds to the fixed group, then unity must or should be one of {d,e,f,g} and not "a".
usually,
e = [0,a,b,c] with a = unity under the seven cycle (a,b,d,c,f,g,e)
g = [0,a,d,e]
c = [0,a,f,g]
now we may only permute the elements {b,c,d,e,f,g} We note that if we simply swap two elements and do so also in the seven cycle itself we will not have the fire smoke and brimstone. We must pay attention to the original four angels that are loosed in the river Euphrates.
Under our original seven cycle we had 'a' as unity for one of the four loosed angels. Thus any seven cycle of the wider set from the "Euphrates" octal;
{[0,c,e,g],[0,c,d,f],[0,b,e,f],[0,b,d,g],[0,a,b,c],[0,a,d,e],[0,a,f,g]}.
would appear to hold fixed [0,c,e,g] corresponding to 'a'=1 and the fixed subgroup complement [0,b,d,f]
Thus by swapping the elements f<=>g under the same original seven cycle we appear to have a=1 and b=>d=>f under frobenius,
thus we are confused with the valid associations of the K4 ultrafilter as thus ...
a = {a,b,c}
b = {a,d,e}
c = {a,f,g}
...but from a different and invalid octal underlying the K4 ultrafilter.
This is a result of preserving the unity element and making an alternate octal from the four angels bound in Euphrates. The original, or "sun" octal is present only in part with three subgroups; (a,b,c),(a,d,e),(a,f,g) here: With a = 1 we have disrupted the octal, yet the K4 ultrafilter is intact. As outside of the K4 ultrafilter, the serpent spirits are synonymous with "Death" as numbered 666. If not one of the four angels are considered "static"  we force the interrupted (or ambiguous) octal and revert to the K4 ultrafilter: (i.e., a=1 and one of (a,b,c),(a,d,e),(a,f,g) must be static.) but this raises some confusion over the underlying octal, after all, if both octals are then valid: Is there an "agreement to disagree"? Not if there is a specific witness in one seven cycle or another to distinguish the two octals. For this we have the famous "Two Witnesses" of revelation.
Likewise from the other three angels we obtain the same, swapping as such elements f<=>g The result is that the operations appear to continue as before, but the "Christ" like part of the ultrafilter is opposed; making the uniqueness of it within the underlying octal confused  its is like "blood" mingling within two different octals.
Before it is said that the serpents mimic Christ, they do not  they supply the four subgroups of the tail (in the constructed alternative octal.)  i.e. their tails have heads and with them they do hurt. They interrupt the octal: although they do not oppose the K4 ultrafilter  If the serpents are to be opposed, then Christ is the one for the task. They, as "Death" or "Wormwood" are overcome in Christ.
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