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Definition S.O.1 - Partial Order(X,E) is a partially ordered set if and only if E c XxX and (1) /\(x:-X) xEx, (2) /\(x,y:-X) xEy and yEx => y = x, (3) /\(x,y,z:-X) xEy and yEz => xEz.Remark:If we write that (X,c) is a partlially ordered set we mean that (X,E) is a partially ordered set where E = {(A,B):-XxX)|A c B}.Definition S.O.2 - Linear Order(X,E) is a linearly ordered set if and only if E c XxX and (1) /\(x:-X) xEx, (2) /\(x,y:-X) xEy and yEx => y = x, (3) /\(x,y,z:-X) xEy and yEz => xEz, (4) /\(x,y:-X) xEy or yEx.Definition S.O.3 - Well Order(X,E) is a well ordered set if and only if E c XxX and (1) /\(x:-X) xEx, (2) /\(x,y:-X) xEy and yEx => y = x, (3) /\(x,y,z:-X) xEy and yEz => xEz, (4) /\(A) ( AcX and A != O => \/(y:-A)/\(z:-A) yEz ).Theorem S.O.4If (X,E) is a linearly ordered set then (X,E) is a partially ordered set.ProofIt is true by Definition S.O.2 and Definition S.O.3.Theorem S.O.5If (X,E) is a well ordered set then (X,E) is a linearly ordered set.ProofTake any a,b:-X. Assume to the contrary that !(aEb) and !(bEa). Let A = {a,b}. Thus /\(z:-A) \/(y:-A) !(zRy). But by Definition S.O.3 we have \/(z:-A)/\(y:-A) zRy. Contradiction. We have proved that aEb or bEa.Definition S.O.6 - Initial SegmentLet (X,E) be a partially ordered set. AcX is an initial segment of X if and only if /\(a:-A)/\(x:-X) xEa => x:-A.Theorem S.O.7If (X,E) is a partially ordered set, KcP(x) and /\(A:-K) A is an initial segment of X then u(K) is an initial segment of X.ProofTake any a:-u(K). Take any x:-X such that xEa. We have A:-K such that a:-A. Since A is an initial segment of X, x:-A. Thus x:-u(K). We have shown that u(K) is an initial segment of X.Theorem S.O.8If (X,E) is a well ordered set, AcX and A != X then A is an initial segment of X if and only if there is a z:-X such that A = {x:-X | xEz and z!=x}.ProofAssume that there is a z:-X such that A = {x:-X | xEz and z!=x}. Take any a:-A. Take any x:-X such that xEa. By transitivity we have xEz and x!=z. Thus x:-A. So A is an initial segment of X. Now, assume that A is an initial segment of X. By Definition S.O.3 we have z:-X\A such that /\(y:-X\A) zEy. It is obvious that {x:-X | xEz and z!=x} c A. We will show that A c {x:-X | xEz and z!=x}. Take any a:-A. Assume to the contrary that zEa. Then by Definition S.O.6 z:-A. Contradiction. We have shown that !(zEa). By Theorem S.O.5 (X,E) is a linearlly ordered set. Thus aEz and z != a. So a:-{x:-X | xEz and z!=x}. We have shown that A = {x:-X | xEz and z!=x}. Now the proof is complete.