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 Th47:
  R is well-ordering & a in field R & b in field R & a <> b
  implies not R |_2 (R-Seg(a)),R |_2 (R-Seg(b)) are_isomorphic
proof
  assume that
A1: R is well-ordering and
A2: a in field R & b in field R and
A3: a <> b;
A4: now
    set S = R |_2 (R-Seg(a));
    assume
A5: R-Seg(b) c= R-Seg(a);
    then
A6: S |_2 (R-Seg(b)) = R |_2 (R-Seg(b)) by Th22;
A7: field S = R-Seg(a) by A1,Th32;
A8: b in R-Seg(a)
    proof
      assume not b in R-Seg(a);
      then not [b,a] in R by A3,Th1;
      then [a,b] in R by A2,A3,A1,Lm4;
      then a in R-Seg(b) by A3,Th1;
      hence contradiction by A5,Th1;
    end;
    then R-Seg(b) = S-Seg(b) by A1,Th27;
    hence thesis by A1,A7,A8,A6,Th25,Th46;
  end;
A9: now
    set S = R |_2 (R-Seg(b));
    assume
A10: R-Seg(a) c= R-Seg(b);
    then
A11: S |_2 (R-Seg(a)) = R |_2 (R-Seg(a)) by Th22;
A12: field S = R-Seg(b) & S is well-ordering by A1,Th25,Th32;
A13: a in R-Seg(b)
    proof
      assume not a in R-Seg(b);
      then not [a,b] in R by A3,Th1;
      then [b,a] in R by A2,A3,A1,Lm4;
      then b in R-Seg(a) by A3,Th1;
      hence contradiction by A10,Th1;
    end;
    then R-Seg(a) = S-Seg(a) by A1,Th27;
    hence thesis by A13,A11,A12,Th40,Th46;
  end;
  R-Seg(a),R-Seg(b) are_c=-comparable by A1,Th26;
  hence thesis by A9,A4;
end;
