. Lemma, Let (X, <) | S be a presentation by operator of A and let F be its reduction family

. Proof, The set Red (F ) is the set of words w such that every sub-word of w belongs to Red (S) From Proposition 2.1.14, a word belongs to Red (S) if and only if it does not belong to lm (R)

<. Assume-that-(-x and . Confluent, Let f ? I(R) We assume that lc (f ) is equal to 1. The kernel of ?F being equal to I(R), (?F ) (f ) is equal to 0. Hence, ?F ) (lm (f )) is equal to (?F ) (lm (f ) ? f )

R. Absence and . Basis, We would like to equip the set RO (G, <) with a lattice structure We cannot use the argument of Section 2.1 because a subspace of KG does not necessarily admit a reduced basis. Indeed, consider G = {g 1 , g 2 , g 3 } ordered such: g 1 < g 3 and g 2 < g 3 . The subspace of KG spanned by g 3 ? g 1 and g 3 ? g 2 does not admit any reduced basis, order to equip RO (G, <) with an order relation, we need the following lemma

. Lemma, Let T 1 and T 2 be two reduction operators relative to (G, <) such that ker (T 1 ) is included in ker, Then, Red (T 2 ) is included in Red

L. Absence and . Structure, The order introduced in 4.1.5 does not induce a lattice structure. Consider G = {g 1 , g 2 , g 3 , g 4 , g 5 } ordered such: g 1 < g 3 , g 1 < g 4 , g 2 < g 3

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