Let (A,<) be a strictly ordered set and b not in A. Define a relation \prec in B=A \cup{b} as follows:
x \precy if and only if (x,y in A and x<y) or (x in A and y=b).
Show that \prec is a strict ordering of B and \prec \cap A^2 =<. (Intuitively, \prec keeps A ordered in the same way as < and makes b greaters than every element of A.)

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