Set Equivalence Theory snippets:
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set1: {pink}={blue}

	proof:

	Let B represent 4 corner 3x3 boxes; they contain 4N

	Let C represent the 4 corners.

	Let E represent the 4 edges. Note that E+C=4N

	Thus, B = E+C  [= 4N]

	Let I represent the intersection of B&E.

	If we delete cells in I,
	then the equality above is preserved, so

		B-I = E-I + C

		blue = E-I + C = pink


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set2: {pink}={blue} (phistomefel)


set12.png:
	obviously pink & blue have equal sets of digits,
	because they have the same deficiency.


set9.png:
	checkerboard...
	4 even cols = 4N
	4 even rows = 4N
	subtract their intersection to get checkerboard pattern
	of 4 red columns + 4 blue rows ...
	whose integer sets match.


set11.png:
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	p': UL-box + (bot+rit)-edges = 3N-99
	b': LR-box + (top+lef)-edges = 3N-11

	so

	p'+99 = b'+11 = 3N

	let i=  p' intersect b'

	pink = p'+99-i

		is still equal to

	blue = b'+11-i
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aads.png:
	Aads set theorem says that the set of digits in the
	pink region is equal to those in the blue region PLUS
	one copy of {1..9} [set called N]

	proof: 
		red+white = 10N-red;
		red+white+blue = 10N-red+blue = 9N
		N-red+blue=0
		red = blue+N

aads2.png: (corollary)
	The same can be said after removing corner blocks.

