, Let w be a factor of length 2 ? . If w can be de-substituted, then we have w = ?(v) for some factor v of length 2 ??1 , and ?(w) = (2|v| 2 , |v| 0 + |v| 1 , |v| 0 + |v| 1 ) Using the inductive hypothesis, it is easy to check that ?(w) = P ? or ?(w) = P ? + (?2, 1, 1) and that the equalities for ?(? ? (0)), ?(? ? (2)) are satisfied. If w cannot be de-substituted, then w occurs at an odd index in x and w is of the form 0, pp.1-1
, 1, 1)}, consider all factors of length 2 ? occurring between two consecutive occurrences of ?(? ? (0)) and ?(? ? (2)). By continuity of the number of 0's, one of these factors must have a Parikh vector equal to
, One can check the first equality for ? = 1. Let ? ? 2 and 2 ??1 ? r ? 2 ? . From the previous lemma, we have max 0 (2 ? + r)
, ? + r) ? min 0 (2 ? ) + min 0 (r) (mod 2) by Lemma A.7. Since we have min 0 (2 ? ) ? 0 (mod 2) by Lemma A
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