theorem Th200:
  a <> 0 implies a ** (A\+\B) = (a**A) \+\ (a**B)
proof
  assume
A1: a <> 0;
  thus a ** (A \+\ B) = (a**(A\B)) \/ (a**(B\A)) by Th92
    .= ((a**A)\(a**B)) \/ (a**(B\A)) by A1,Th199
    .= (a**A) \+\ (a**B) by A1,Th199;
end;
