[1Z7] (Replaces 06G) (Replaces 06H) Consider \(R\) a transitive and reflexive relation in \(A× A\); such a relation is called a preorder [ 42 ] ; we define \(x ∼ y\iff (xRy ∧ yRx)\) then \(∼\) is an equivalence relation, \(R\) is invariant for \(∼\), and \(\widetilde R\) (defined as in [1Z6]) is an order relation.
\(∼\) is clearly reflexive and symmetrical; is transitive because if \(x∼ y,y∼ z\) then \(xRy∧ yRx∧ yRz∧ zRy\) but being \(R\) transitive you get \(xRz∧ zRx\) i.e. \(x∼ z\)
Let \(x,y,\tilde x,\tilde y∈ X\) be such that \(x∼ \tilde x,y∼ \tilde y\) then we have \(xR\tilde x ∧ \tilde xRx∧ yR\tilde y∧ \tilde yRy\) if we add \(xRy\), by transitivity we get \(\tilde xR\tilde y\); and symmetrically.
Finally, we see that \({\widetilde R}\) is an order relation on \(Y\). Using the (well posed) definition ”\([x]{\widetilde R}[y] \iff xRy\)” we deduce that \({\widetilde R}\) is reflexive and transitive (as indeed stated in the previous proposition). \({\widetilde R}\) is also antisymmetric because if for \(z,w∈ A/∼\) you have \(z{\widetilde R}w∧ w{\widetilde R}z\) then taken \(x∈ z,y∈ w\) we have \(xRy ∧ yRx\) which means \(x∼ y\) and therefore \(z=w\).