theorem Th47:
  k <> {} implies (i/k)+(j/k) = (i+^j)/k
proof
  assume
A1: k <> {};
  hence (i/k)+(j/k) = (i*^k+^j*^k)/(k*^k) by Th46
    .= ((i+^j)*^k)/(k*^k) by ORDINAL3:46
    .= (i+^j)/k by A1,Th44;
end;
