Stirring Up His Power And Courage

The king of the north atacks the south again, we see the collision of the north and south play out in the text - the king of the north has done his preparation - and there is tested the structure of the octal between the left and right hands we would see in Christ.

-- Click To Expand/Collapse Bible Verses -- Dan ch11:v25-26
Dan 11:25 And he shall stir up his power and his courage against the king of the south with a great army; and the king of the south shall be stirred up to battle with a very great and mighty army; but he shall not stand: for they shall forecast devices against him.
Dan 11:26 Yea, they that feed of the portion of his meat shall destroy him, and his army shall overflow: and many shall fall down slain.

Stirring up his power (the static subgroup [b,d,f]) and his courage (the singletons a, c, e and g) against the king of the south we see that the king of the north is now positioned in the southern octal prepared for battle.

The king of thee south likewise prepares in a symmetrical way, but does not stand becuase the north "forecast devices" against him.

Stirring up his power
b = [c,d,g]
d = [c,e,f]
f = [b,e,g]

and his courage
a = [b,d,f]
c = [a,f,g]
e = [a,b,c]
g = [a,d,e]

against the king of the south that has in respect the octal triples (a mighty army stirred up)
b = {a,e,f}
d = {b,c,g}
f = {a,c,d}
a ={c,e,g}
c = {b,d,e}
e = {d,f,g}
g = {b,c,f}

But the south does not stand because the devices against him fromo the north depend on a seven cycle. (By not "standing" we refer to the situation that the south is strictly limited to static triples (bows) rather than subgroups, (the "power of the arm").

By "forecasting devices" we have that for the static subgroup (power) of the north and the "courage" of the central element "in battle" teh seven cycles that permit this situation are such as for [b,d,f] static

(a,b,d,c,f,g,e) and (a,b,f,c,d,e,g) correspond to a = 1
(c,b,d,a,f,e,g) and (c,b,f,a,d,g,e) correspond to c = 1
(e,b,d,g,f,c,a) and (e,b,f,g,d,a,c) correspond to e = 1
(g,b,d,e,f,a,c) and (g,b,f,e,d,c,a) correspond to g = 1

Then for each seven cycle we have the situation (using the top row of cycles)

And we see those that eat of the portion of his meat (those that sit at his "table" with him) destroy him. Those singletons c, e, or g in the centre are opposed by subgroups forming a K4 group that have the element "a" for their intersection. Within the fortress of the southern octal are elements from the static subgroup of the king of the north corresponding to his army - the element "a". The elements then in the southern fortress opposing the devices do not form a K4 group. The north has won, and has reduced the structure of the south to remain "less rich" than the north. (The south is "destroyed" - and can not form the same devices.)

The king's army "overflows" - (ie that of the north) so that we have the northern fortress [b,d,f] common to the four possible cosets with a, c, e or g = 1. This requires a change of seven cycle - but for each seven cycle used this forecasting of devices indeed holds - and the king of the north fully succeeds against the south.

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