theorem Th29:
  a > 0 implies a to_power (b-c) = a to_power b / a to_power c
proof
  assume
A1: a > 0;
then  a #R (b-c) = a #R b / a #R c by PREPOWER:77;
then  a #R (b-c) = a #R b / a to_power c by A1,Def2;
then  a #R (b-c) = a to_power b / a to_power c by A1,Def2;
  hence thesis by A1,Def2;
end;
