reserve i, j, n for Element of NAT,
  f, g, h, k for FinSequence of REAL,
  M, N for non empty MetrSpace;

theorem Th15:
  len f = len h & len g = len k implies abs (f^g - h^k) = abs (f-h
  ) ^ abs (g-k)
proof
  assume that
A1: len f = len h and
A2: len g = len k;
A3: len (f^g) = len f + len g by FINSEQ_1:22;
A4: len (h^k) = len h + len k by FINSEQ_1:22;
A5: len abs (f^g - h^k) = len (f^g - h^k) by Th9
    .= len (f^g) by A1,A2,A3,A4,Th7
    .= len (f-h) + len g by A1,A3,Th7
    .= len (f-h) + len (g-k) by A2,Th7
    .= len abs (f-h) + len (g-k) by Th9
    .= len abs (f-h) + len abs (g-k) by Th9
    .= len (abs (f-h) ^ abs (g-k)) by FINSEQ_1:22;
  for i be Nat st 1 <= i & i <= len abs (f^g - h^k) holds (abs (f^g - h^k)
  ).i = (abs (f-h) ^ abs (g-k)).i
  proof
    let i be Nat;
    assume that
A6: 1 <= i and
A7: i <= len abs (f^g - h^k);
A8: i in dom abs (f^g - h^k) by A6,A7,FINSEQ_3:25;
    then
A9: i in dom (f^g - h^k) by Th9;
    per cases;
    suppose
A10:  i in dom f;
      reconsider i1 = i as Element of NAT by ORDINAL1:def 12;
A11:  i in dom h by A1,A10,FINSEQ_3:29;
A12:  i in dom (f-h) by A1,A10,Th7;
      then
A13:  i in dom abs (f-h) by Th9;
      thus (abs (f^g - h^k)).i = |.(f^g - h^k).i1.| by A8,TOPREAL6:12
        .= |.(f^g).i - (h^k).i.| by A9,VALUED_1:13
        .= |.f.i - (h^k).i.| by A10,FINSEQ_1:def 7
        .= |.f.i - h.i.| by A11,FINSEQ_1:def 7
        .= |.(f-h).i1.| by A12,VALUED_1:13
        .= (abs (f-h)).i1 by A13,TOPREAL6:12
        .= (abs (f-h) ^ abs (g-k)).i by A13,FINSEQ_1:def 7;
    end;
    suppose
A14:  not i in dom f;
      reconsider i1 = i as Element of NAT by ORDINAL1:def 12;
A15:  len f < i by A6,A14,FINSEQ_3:25;
      then reconsider j = i - len f as Element of NAT by INT_1:5;
A16:  len (f-h) < i by A1,A15,Th7;
      then
A17:  len abs (f-h) < i by Th9;
      i <= len (f^g - h^k) by A7,Th9;
      then
A18:  i <= len (f^g) by A1,A2,A3,A4,Th7;
      then i <= len (f-h) + len g by A1,A3,Th7;
      then i <= len (f-h) + len (g-k) by A2,Th7;
      then i - len (f-h) in dom (g-k) by A16,Th4;
      then
A19:  j in dom (g-k) by A1,Th7;
      then
A20:  j in dom abs (g-k) by Th9;
      len f = len (f-h) by A1,Th7;
      then
A21:  j = i - len abs (f-h) by Th9;
      thus (abs (f^g - h^k)).i = |.(f^g - h^k).i1.| by A8,TOPREAL6:12
        .= |.(f^g).i - (h^k).i.| by A9,VALUED_1:13
        .= |.g.j - (h^k).i.| by A15,A18,FINSEQ_1:24
        .= |.g.j - k.j.| by A1,A2,A3,A4,A15,A18,FINSEQ_1:24
        .= |.(g-k).j.| by A19,VALUED_1:13
        .= (abs (g-k)).j by A20,TOPREAL6:12
        .= (abs (f-h) ^ abs (g-k)).i by A5,A7,A17,A21,FINSEQ_1:24;
    end;
  end;
  hence thesis by A5,FINSEQ_1:14;
end;
