theorem Th32:
  a>0 implies (1/a) to_power b = a to_power (-b)
proof
  assume
A1: a>0;
  hence (1/a) to_power b = (1/a) #R b by Def2
    .= 1/a #R b by A1,PREPOWER:79
    .= 1/a to_power b by A1,Def2
    .= a to_power (-b) by A1,Th28;
end;
