
theorem Th2:
  for R being non empty 1-sorted,
      X being Subset of R holds
    { x where x is Element of R : {} c= X } = [#]R
  proof
    let R be non empty 1-sorted;
    let X be Subset of R;
    thus { x where x is Element of R : {} c= X } c= [#]R
    proof
      let y be object;
      assume y in { x where x is Element of R : {} c= X }; then
      ex z being Element of R st z = y & {} c= X;
      hence thesis;
    end;
    let y be object;
    assume
A1: y in [#]R;
    y in { x where x is Element of R : {} c= X}
    proof
      assume not y in { x where x is Element of R : {} c= X }; then
      not y is Element of R or not {} c= X;
      hence contradiction by A1;
    end;
    hence thesis;
  end;
