theorem
  a * Partial_Sums(seq1) + b * Partial_Sums(seq2) = Partial_Sums(a *
  seq1 + b * seq2)
proof
  thus a * Partial_Sums(seq1) + b * Partial_Sums(seq2) = Partial_Sums(a * seq1
  ) + b * Partial_Sums(seq2) by Th3
    .= Partial_Sums(a * seq1) + Partial_Sums(b * seq2) by Th3
    .= Partial_Sums(a * seq1 + b * seq2) by Th1;
end;
