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 Metamath Wormwood = Death We constructed the third part of a "quasi product" upon a relation between different octals (in the same seven symbols) namely using as previously "fire columns" as follows. Take an octal with a seven cycle and form the left handed (fire) column, e.g. using (a,b,d,c,f,g,e) a = (b,d,f) = (c,e,g) * b = (c,d,g) = (a,e,f) d = (c,e,f) = (a,b,g) c = (a,f,g) = (b,d,e) f = (b,e,g) = (a,c,d) g = (a,d,e) = (b,c,f) e = (a,b,c) = (d,f,g) And now repeat using any other seven cycle on the same octal. ie (a,b,f,c,d,e,g) a = (b,d,f) = (c,e,g) * b = (c,e,f) = (a,d,g) f = (c,d,g) = (a,b,e) c = (a,d,e) = (b,f,g) d = (b,e,g) = (a,c,f) e = (a,f,g) = (b,c,d) g = (a,b,c) = (d,e,f) Notice the shared pair marked by the asterisk Now take another seven cycle on the original octal say using (a,c,e,b,f,g,d) a = (c,e,f) = (b,d,g) c = (b,e,g) = (a,d,f) e = (b,d,f) = (a,c,g) b = (a,f,g) = (c,d,e) f = (c,d,g) = (a,b,e) \$ g = (a,d,e) = (b,c,f) % d = (a,b,c) = (e,f,g) Now the pairs marked with % and \$ are present also in the first two sets, one from each. Now we note that there is an easily determined seven cycle on an octal that has the same marked pairs - ie using (a,d,b,e,f,c,g) a = (b,d,f) = (c,e,g) * f = (c,d,g) = (a,b,e) \$ g = (a,d,e) = (b,c,f) % which is in whole form a = (b,d,f) = (c,e,g) * d = (b,c,e) = (a,f,g) b = (e,f,g) = (a,c,d) e = (a,c,f) = (b,d,g) f = (c,g,d) = (a,b,e) \$ c = (a,b,g) = (d,e,f) g = (a,d,e) = (b,c,f) % Which shares one of each of the associations. This is our "wormwood" set. (just to demonstrate we can generate a different octal from a common one) How many of these uniquely determined wormwood sets are there? We started with eight choices of C7 groups on our original octal, then another seven choices, then a final six which determined the "wormwood" set. (8x7x6 = 336). We note that the wormwood set is defined by two columns (octals), neither of which are capable of being our original octal (unless the same seven cycle is used in our choices (we excluded this possibility). Then in the original case, we would have that all three candidates/correspondences are subgroups of the original octal, not the two we require from each of the three to make the wormwood set. So we may multiply by these two possible octals, in the knowledge we are not including the original octal or its complement, so 336 x 2 = 672. Then we note that there is the possibility that these three choices actually do have an intersection common (as "d" in the middle column of the wormwood set) These indeed form a group of K4 subgroups in their octal as here in the middle column. Were there the possibility of choosing three elements as a group of subgroups in our original octal, say (a,d,e),(a,f,g),(a,b,c) they would also have a common intersection ("a" here). We must yet again exclude the possibility of our original octal. Here in this particular case we have a = (b,d,f) = (c,e,g) * f = (c,d,g) = (a,b,e) \$ g = (a,d,e) = (b,c,f) % Yet we must not include (a,f,g) as a subgroup of the middle column octal in the wormwood set, but in the rightmost column instead where the alignment to the singletons on the left column occurs. There are six octals that contain (a,f,g) as a subgroup - as for the middle column we can not generate those six octals (lest we generate our original octal as our wormwood set) - those very six ones containing (a,f,g). This means that six octals out of all those (336x2) may not or can not be generated. (although (a,f,g) would be within the rightmost column as above). So noting that six octals are excluded from one column (in the middle here), we must subtract six from the 672. (and not twelve.) We now have 666. Our three choices of correspondence *, % and \$ must (in each middle and right column) never make a subgroup in the original octal. We then may repeat the problem of there being "bows" as with the fourth seal problem. We have a direct analogy,.. though here we are using the seven cycles to show there is a similarity. Instead of choosing different octals we are choosing different seven cycles - however in the knowledge that for each bow held fixed there is an equivalent seven cycle - we actually have a 1-1 coorrespondence between the number 666 upon the octals and the seven cycles as shown here. Wormwood is in fact, the name of "Death" - the only horseman of the seals of revelation. We have simply used the analogy of shifting "bows" from the trumpets rather than shifting a choice of identity as done in the first four seals. Since we can choose the number of "bows" that form a "quad", we have the one "Death" bow or wormwood set aligns with the three corresponding bows (as the first three seals) or "correspondences from seven cycles" as above. We note that in the wormwood octal with those three correspondences we may simply choose "d" = 1 with those correspondences from the first three sets and we have the equivalent problem to the "666 outside of Christ" numbering to 666 shown before. Rev 8:11 And the name of the star is called Wormwood: and the third part of the waters became wormwood; and many men died of the waters, because they were made bitter. The revelation shows that "many" died of bitter water - the Holy Spirit is the living water given by Christ, here we simply see the evidence of satan's counterfeit - bitter waters that kill. Christ is life, antichrist is Death. We also see from the octal that the ultrafilter is interrupted as described in the sixth seal account later in revelation. (with a transposition causing Death with fire smoke and brimstone - see elsewhere). Likewise we also now have the simple answer that Abbadon or Apollyon is "Hell" that follows after. (see fifth trumpet.) This occurs with the setting up of the image to the beast.Continue To Next Page Return To Section Start Return To Previous Page