reserve a,b,c,d,x,y,z for object, X,Y,Z for set;
reserve R,S,T for Relation;
reserve F,G for Function;

theorem Th29:
  R is well-ordering & a in field R & b in field R implies ( [a,b]
  in R iff R-Seg(a) c= R-Seg(b) )
proof
  assume that
A1: R is well-ordering and
A2: a in field R and
A3: b in field R;
  thus [a,b] in R implies R-Seg(a) c= R-Seg(b)
  proof
    assume
A4: [a,b] in R;
    let c be object;
    assume
A5: c in R-Seg(a);
    then
A6: [c,a] in R by Th1;
    then
A7: [c,b] in R by A1,A4,Lm2;
    c <> a by A5,Th1;
    then c <> b by A1,A4,A6,Lm3;
    hence thesis by A7,Th1;
  end;
  assume
A8: R-Seg(a) c= R-Seg(b);
   now
    assume
A9: a <> b;
    assume not [a,b] in R;
    then [b,a] in R by A2,A3,A1,A9,Lm4;
    then b in R-Seg(a) by A9,Th1;
    hence contradiction by A8,Th1;
  end;
  hence thesis by A1,A2,Lm1;
end;
