a. 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 iff (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 < makes b greater than every element of A.)

b. Generalize part (a): Let ( A_1, <_1) and ( A_2, <_2) be strict orderings, A_1 \cap A_2 = \emptyset. Define a relation \prec on B =  A_1 \cup A_2 as follows:
x \precy iff x,y in A_1 and  x <_1 y
or x,y in A_2 and  x <_2 y
or x in A_1 and y in A_2

Show that \prec is a strict ordering of B and \prec \cap A^{2}_1 = <_1, \prec \cap A^{2}_2=<_2. (Intuitively, \prec puts every element of A_1 before every element of A_2 and coincides with the original orderings of A_1 and A_2

Does anyone have any suggestions on how to do this?