theorem Th82:
  len p = k + 1 & q = p | Seg k implies (not p.(k + 1) in A iff p
  - A = (q - A) ^ <* p.(k + 1) *>)
proof
  assume that
A1: len p = k + 1 and
A2: q = p | Seg k;
  thus not p.(k + 1) in A implies p - A = (q - A) ^ <* p.(k + 1) *>
  proof
    assume
A3: not p.(k + 1) in A;
    thus p - A = (q ^ <* p.(k + 1) *>) - A by A1,A2,Th53
      .= (q - A) ^ (<* p.(k + 1) *> - A) by Lm11
      .= (q - A) ^ <* p.(k + 1) *> by A3,Lm6;
  end;
  assume
A4: p - A = (q - A) ^ <* p.(k + 1) *>;
  assume p.(k + 1) in A;
  then q - A = (q - A) ^ <* p.(k + 1) *> by A1,A2,A4,Th81;
  hence contradiction by FINSEQ_1:87;
end;
