User:Inthar/MV3: Difference between revisions

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 then we (can assume we) have one of T1+X, T2+X, or T3+X. Scoot this left and you lose the X on the right, and gain another non-X letter on the left so you get a fourth variant of this interval class that contains T1 + X, a contradiction.
 === Sizes of chunks of any fixed letter form a MOS ===
+WOLOG consider chunks of X. Use Q for both Y and Z.
+First we prove that chunk sizes can't differ by 2 or more.
+By elimination we can assume that we have at least three Qs.
+have some length (say that of max chunk) word with no q's,one q and 2 q's.
+now y[Max]z => contradiction bc two kinds of "one q"
+so some non Max chunk has to have y[chunk]z (or z...y)
+then by using size of [y[non-max chunk]z]  you get a contradiction bc you can scoot to get an x (since consecutive q's cant happen if there are consecutive x's)
+u get [xyxxx...x]z, x[yxxxx...xz]x, y[x...xz], and all x's from the max chunk
+If we have more q's than x's then we can't have "xx", so we're done.
+This proves the claim about chunk sizes.