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On systems of word equations with simple loop sets

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then u 1 and u 2 commute.<br />

Suppose<br />

x 0 u<br />

k+1<br />

1<br />

x 0 u<br />

k+1<br />

1<br />

v k+1<br />

1<br />

u 1<br />

u 2<br />

|v k+1<br />

1 | < |x 0 | + |u k+2<br />

1 |.<br />

This implies, together <strong>with</strong> (11), that |v k 1| < |x 0 u 1 |. Let d = v −k<br />

1 x 0 u 1 . Note<br />

that d is a prefix <strong>of</strong> v 1 .<br />

If y 1 ≠ ǫ then v 1 and y 1 are not marked. Therefore y 1 = ǫ and d is comparable<br />

<strong>with</strong> v 2 . If |d| ≥ |v 2 | then v 1 and v 2 commute.<br />

d<br />

<br />

x 0 u 1<br />

v k 1<br />

v 1<br />

v k 1<br />

v 2<br />

Suppose the contrary, which implies that d is a prefix <strong>of</strong> v 2 (as well as <strong>of</strong> v 1 ).<br />

Then both x 0 u k+2<br />

1 and x 0 u k+1<br />

1 u 2 are comparable <strong>with</strong> v1 k+1 d. Since<br />

|v k+1<br />

1 d| = |x 0 u 1 | + |v 1 | > |x 0 u k+2<br />

1 |,<br />

the <strong>word</strong>s u 1 and u 2 are comparable, and therefore commute.<br />

x 0 u<br />

k+1<br />

1<br />

x 0 u<br />

k+1<br />

1<br />

v k 1<br />

v 1<br />

u 1<br />

u 2<br />

d<br />

Suppose then that the second inequality <strong>of</strong> (7) is not true, that is |v k+1<br />

1 | <<br />

|x 0 u 1 | − 1. Then either v 1 and y 1 are not marked, or (if y 1 = ǫ) the <strong>word</strong>s v 1<br />

and v 2 commute.<br />

x 0 u 1<br />

v k 1<br />

v 1<br />

v k 1<br />

y 1 v 2<br />

The pro<strong>of</strong> is now complete.<br />

We make use <strong>of</strong> the previous theorem when proving our second main result.<br />

8

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