Contradict the Minimality of < B 2 ; C 2 >. If B 3 < C 0 Acknowledgement: I Would like to Thank Ralph Mckenzie for Suggesting That

2 c 2 , then the same argument shows c 0 2 < b 2. Thus b 3 c 2 .Let c 3 be another element of D a covered by c 2. If f(c 3 ; b 3) = b 3 , we have f(c 3 ; b 2) b 3 and f(c 3 ; b 2) c 2 , thus f(c 3 ; b 2) 2 fb 3 ; c 2 g; but this is impossible. Therefore f(c 3 ; b 3) = c 3 , and b 3 > d, since otherwise we would have b 3 = d, and by (4), f(c 3 ; b 3) = d 6 = c 3. Now f(b 3 ; c 3) f(b 3 ; c 2) = b 3 implies f(b 3 ; c 3) = b 3. By the minimality of < b 2 , c 2 >, c 3 must be incomparable to d. But by (4) this is impossible since d = f(c 3 ; d) f(c 3 ; b 3) = c 3. 2