
theorem Th1:
  for a being Element of NAT, q being Real st 1 < q & q |^ a = 1 holds a = 0
proof
  let a be Element of NAT, q be Real such that
A1: 1 < q and
A2: (q |^ a) = 1 and
A3: a <> 0;
  a < 1 + 1 by A1,A2,PREPOWER:13;
  then a <= 0 + 1 by NAT_1:13;
  then a = 1 by A3,NAT_1:8;
  then (q #Z a) = q by PREPOWER:35;
  hence contradiction by A1,A2,PREPOWER:36;
end;
