Discrete Algebraic Methods

Proof: Since s ∼J t, we can write s = ptu and t = qsυ.

(a)Since s ≤R t, we may assume that s = tu ∈ tM. Hence,

This implies t ≤R s. We conclude s ∼R t.

(b)We have

As a consequence, s ≤L qs ≤L s and hence, s ∼L qs. By symmetry; s ∼R sυ. This implies qsM = qsυM = tM; and qs ∈ L(s) ∩ R(t).

(c)We have shown that L(s) ∩ R(t) ≠ 0. Thus, there exists z with s ∼L z ∼R t. We obtain J ⊆ L ∘ R. The other inclusion L ∘ R ⊆ J is trivial. Therefore, J = L ∘ R. By symmetry, J = R∘L. By definition,D = L∘R. Thisproves the claim J = D = L∘R = R∘L. Now, D is reflexive ...