theorem Th5:
  not o on A & not o on B implies rng IncProj(A,o,B) = CHAIN(B)
proof
  assume
A1: ( not o on A)& not o on B;
A2: now
    let x be object;
    assume
A3: x in CHAIN(B);
    then reconsider x9 = x as Element of the Points of IPP;
    ex b st b=x & b on B by A3;
    then consider y such that
A4: y on A and
A5: IncProj(A,o,B).y = x9 by A1,Th3;
    y in CHAIN(A) by A4;
    then y in dom IncProj (A,o,B) by A1,Th4;
    hence x in rng IncProj(A,o,B) by A5,FUNCT_1:def 3;
  end;
  now
    let x be object such that
A6: x in rng IncProj(A,o,B);
    rng IncProj(A,o,B) c= the Points of IPP by RELAT_1:def 19;
    then reconsider x9 = x as POINT of IPP by A6;
    x9 on B by A1,A6,PROJRED1:21;
    hence x in CHAIN(B);
  end;
  hence thesis by A2,TARSKI:2;
end;
