theorem
  for HPS being satisfying_euclidean Heptatonic_Pythagorean_Score
  for frequency being Element of HPS holds
  [heptatonic_pythagorean_scale(HPS,frequency).5,
    heptatonic_pythagorean_scale_ascending(HPS,frequency).2]
  in fifth(HPS) &
  [heptatonic_pythagorean_scale(HPS,frequency).6,
    heptatonic_pythagorean_scale_ascending(HPS,frequency).3]
  in fifth(HPS) &
  not [heptatonic_pythagorean_scale(HPS,frequency).7,
    heptatonic_pythagorean_scale_ascending(HPS,frequency).4]
  in fifth(HPS)
  proof
    let HPS be satisfying_euclidean Heptatonic_Pythagorean_Score;
    let frequency be Element of HPS;
A1: intrval(hepta_4(HPS,frequency),hepta_1(HPS,Octave(HPS,frequency)))
      = (3 qua Real) / 2 &
    intrval(hepta_5(HPS,frequency),hepta_2(HPS,Octave(HPS,frequency)))
      = (3 qua Real) / 2 &
    intrval(hepta_6(HPS,frequency),hepta_3(HPS,Octave(HPS,frequency)))
      <> (3 qua Real) / 2 &
    intrval(hepta_octave(HPS,frequency),hepta_4(HPS,Octave(HPS,frequency)))
      = (3 qua Real) / 2 by Th92;
    reconsider n2 = 2,n3 = 3 as Element of HPS by Th20;
    (the Ratio of HPS).(n2,n3) = @n3 / @n2 by Def25
                              .= (3 qua Real) / 2;
    then (the Ratio of HPS).(hepta_4(HPS,frequency),
      hepta_1(HPS,Octave(HPS,frequency)))
      = (the Ratio of HPS).(n2,n3) &
         (the Ratio of HPS).(hepta_5(HPS,frequency),
      hepta_2(HPS,Octave(HPS,frequency)))
      = (the Ratio of HPS).(n2,n3) &
         (the Ratio of HPS).(hepta_6(HPS,frequency),
      hepta_3(HPS,Octave(HPS,frequency)))
      <> (the Ratio of HPS).(n2,n3) by A1,Th93;
    then n2,n3 equiv hepta_4(HPS,frequency),
      hepta_1(HPS,Octave(HPS,frequency)) &
      n2,n3 equiv hepta_5(HPS,frequency),hepta_2(HPS,Octave(HPS,frequency)) &
      not n2,n3 equiv hepta_6(HPS,frequency),hepta_3(HPS,Octave(HPS,frequency))
        by Def08a;
    hence thesis by EQREL_1:18;
  end;
