By Inverse Completion is Commutative Semigroup: :every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$. === Proof that Relation is Well-Defined === First we need to show that $\preceq'$ is well-defined. So we need to show that if $x_1, x_2, y_1, y_2, z_1, z_2, w_1, w_2 \in S$ satisfy: {{begin-eqn}} {{eqn | l = x_1 \circ \paren {y_1}^{-1}       | r = x_2 \circ \paren {y_2}^{-1}       | c =  }} {{eqn | l = z_1 \circ \paren {w_1}^{-1}       | r = z_2 \circ \paren {w_2}^{-1}       | c =  }} {{eqn | l = x_1 \circ w_1       | o = \preceq       | r = y_1 \circ z_1       | c =  }} {{end-eqn}} then: :$x_2 \circ \paren {w_2}^{-1} = y_2 \circ \paren {z_2}^{-1}$ We have: {{begin-eqn}} {{eqn | l = x_1 \circ \paren {y_1}^{-1}       | r = x_2 \circ \paren {y_2}^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x_1 \circ y_2       | r = x_2 \circ y_1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = z_1 \circ \paren {w_1}^{-1}       | r = z_2 \circ \paren {w_2}^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = z_1 \circ w_2       | r = z_2 \circ w_1       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = x_2 \circ w_2 \circ y_1 \circ z_1       | r = x_1 \circ w_1 \circ y_2 \circ z_2       | c =  }} {{eqn | o = \preceq       | r = y_1 \circ z_1 \circ y_2 \circ z_2       | c =  }} {{eqn | ll= \leadsto       | l = x_2 \circ w_2       | o = \preceq       | r = y_2 \circ z_2       | c = Strict Ordering Preserved under Product with Cancellable Element }} {{end-eqn}} Thus $\preceq'$ is a well-defined relation on $T$. {{qed|lemma}} === Proof that Relation is Transitive === To show that $\preceq'$ is transitive: {{begin-eqn}} {{eqn | ll= \forall x, y, z, w, u, v \in S:       | l = x \circ y^{-1}       | o = \preceq'       | r = z \circ w^{-1}       | c =  }} {{eqn | l = z \circ w^{-1}       | o = \preceq'       | r = u \circ v^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x \circ w \circ v       | o = \preceq       | r = y \circ z \circ v       | c =  }} {{eqn | o = \preceq       | r = y \circ w \circ u       | c =  }} {{eqn | ll= \leadsto       | l = x \circ v       | o = \preceq       | r = y \circ u       | c = Strict Ordering Preserved under Product with Cancellable Element }} {{eqn | ll= \leadsto       | l = x \circ y^{-1}       | o = \preceq'       | r = u \circ v^{-1}       | c =  }} {{end-eqn}} Thus $\preceq'$ is transitive. {{qed|lemma}} === Proof that Relation is Total Ordering === To show that $\preceq'$ is a total ordering on $T$ compatible with $\circ$: Let $z_1, z_2 \in T$. Then: :$\exists x_1, x_2, y_1, y_2 \in S: z_1 = x_1 \circ y_1^{-1}, z_2 = x_2 \circ y_2^{-1}$ Then: :$z_1 \circ y_1 = x_1$ :$z_2 \circ y_2 = x_2$ As $\preceq$ is a total ordering on $S$, either: :$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$ or: :$z_2 \circ y_2 \circ y_1 \preceq z_1 \circ y_1 \circ y_2$ {{WLOG}}, suppose that: :$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$ So: {{begin-eqn}} {{eqn | l = z_1 \circ y_1 \circ y_2       | o = \preceq       | r = z_2 \circ y_2 \circ y_1       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_2       | o = \preceq       | r = x_2 \circ y_1       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_1^{-1}       | o = \preceq'       | r = x_2 \circ y_2^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = z_1       | o = \preceq'       | r = z_2       | c =  }} {{end-eqn}} and it is seen that $\preceq'$ is a total ordering on $T$. {{qed|lemma}} === Proof that Relation is Compatible with Operation === Let $x_1, x_2, y_1, y_2 \in T$ such that $x_1 \preceq' x_2, y_1 \preceq' y_2$. We need to show that $x_1 \circ y_1 \preceq' x_2 \circ y_2$. Let: : $x_1 = r_1 \circ s_1^{-1}, x_2 = r_2 \circ s_2^{-1}$ : $y_1 = u_1 \circ v_1^{-1}, y_2 = u_2 \circ v_2^{-1}$ We have: {{begin-eqn}} {{eqn | l = x_1       | o = \preceq'       | r = x_2       | c =  }} {{eqn | ll= \leadsto       | l = r_1 \circ s_1^{-1}       | o = \preceq'       | r = r_2 \circ s_2^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = r_1 \circ s_2       | o = \preceq       | r = r_2 \circ s_1       | c =  }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = y_1       | o = \preceq'       | r = y_2       | c =  }} {{eqn | ll= \leadsto       | l = u_1 \circ v_1^{-1}       | o = \preceq'       | r = u_2 \circ v_2^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = u_1 \circ v_2       | o = \preceq       | r = u_2 \circ v_1       | c =  }} {{end-eqn}} Because of the compatibility of $\preceq$ with $\circ$ on $S$: {{begin-eqn}} {{eqn | l = \paren {r_1 \circ s_2} \circ \paren {u_1 \circ v_2}       | o = \preceq       | r = \paren {r_2 \circ s_1} \circ \paren {u_2 \circ v_1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {r_1 \circ u_1} \circ \paren {s_2 \circ v_2}       | o = \preceq'       | r = \paren {r_2 \circ u_2} \circ \paren {s_1 \circ v_1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {r_1 \circ u_1} \circ \paren {s_1 \circ v_1}^{-1}       | o = \preceq'       | r = \paren {r_2 \circ u_2} \circ \paren {s_2 \circ v_2}^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {r_1 \circ s_1^{-1} } \circ \paren {u_1 \circ v_1^{-1} }       | o = \preceq'       | r = \paren {r_2 \circ s_2^{-1} } \circ \paren {u_2 \circ v_2^{-1} }       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_1       | o = \preceq'       | r = x_2 \circ y_2       | c =  }} {{end-eqn}} Thus compatibility is proved. {{qed|lemma}} === Proof about Restriction of Relation === To show that the restriction of $\preceq'$ to $S$ is $\preceq$: {{begin-eqn}} {{eqn | l = \forall x, y \in S: x       | o = \preceq       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \forall a \in S: \paren {x \circ a} \circ a       | o = \preceq       | r = \paren {y \circ a} \circ a       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \paren {x \circ a} \circ a^{-1}       | c =  }} {{eqn | o = \preceq       | r = \paren {y \circ a} \circ a^{-1}       | c =  }} {{eqn | r = y       | c =  }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = \forall x, y \in S: x       | o = \preceq'       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \exists u, v, z, w \in S: x       | r = u \circ v^{-1}       | c =  }} {{eqn | l = y       | r = z \circ w^{-1}       | c =  }} {{eqn | l = u \circ w       | o = \preceq       | r = z \circ v       | c =  }} {{eqn | ll= \leadsto       | l = x \circ v \circ w       | r = u \circ w       | c =  }} {{eqn | o = \preceq       | r = z \circ v       | c =  }} {{eqn | r = y \circ v \circ w       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \preceq       | r = y       | c = Strict Ordering Preserved under Product with Cancellable Element }} {{end-eqn}} {{qed|lemma}} === Proof that Relation is Unique === To show that $\preceq'$ is unique: Let $\preceq^*$ be any ordering on $T$ compatible with $\circ$ that induces $\preceq$ on $S$. Then: {{begin-eqn}} {{eqn | ll= \forall x, y, z, w \in S:       | l = x \circ y^{-1}       | o = \preceq^*       | r = z \circ w^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ w       | o = \preceq       | r = y \circ z       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x \circ y^{-1}       | o = \preceq^*       | r = z \circ w^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x \circ w       | r = \paren {x \circ y^{-1} } \circ \paren {y \circ w}       | c =  }} {{eqn | o = \preceq       | r = \paren {z \circ w^{-1} } \circ \paren {y \circ w}       | c =  }} {{eqn | r = y \circ z       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x \circ w       | o = \preceq       | r = y \circ z       | c =  }} {{eqn | ll= \leadsto       | l = x \circ y^{-1}       | r = \paren {x \circ w} \circ w^{-1} \circ y^{-1}       | c =  }} {{eqn | o = \preceq^*       | r = \paren {y \circ z} \circ w^{-1} \circ y^{-1}       | c =  }} {{eqn | r = z \circ w^{-1}       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | ll= \forall x, y, z, w \in S:       | l = x \circ y^{-1}       | o = \preceq^*       | r = z \circ w^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ w       | o = \preceq       | r = y \circ z       | c =  }} {{end-eqn}} Hence as every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$, the orderings $\preceq^*$ and $\preceq'$ are identical. {{qed}}