S(R)=R ∪ R (inverse) = {
S(R)=R( 1 order) ∪ r (order 2) ∪ r (order 3) ... = {< a, b>,<b, a>,<b, c>},<c, b>.
R( 1 order) = r = {
R (second order) = {
R (third order) = {
R (fourth order) = {
Therefore: s (r) = {
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