reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r,s for Real;
reserve p,p1,p2,p3 for Prime;

theorem Th100:
  n > 1 implies 3|^n - 3 to_power(1-n) - 2 > 0
  proof
    assume
A1: n > 1;
    set a = 3 to_power n;
A2: 3 to_power(1-n) = 3 to_power 1 / 3 to_power n by POWER:29
    .= 3 / a;
A3: a*3/a = 3 by XCMPLX_1:89;
A4: a*(a-3/a-2) = a*a-a*3/a-2*a
    .= (a+1)*(a-3) by A3;
    now
      assume a-3/a-2 <= 0;
      then a-3 <= 0 by A4;
      then a-3+3 <= 0+3|^1 by XREAL_1:6;
      hence contradiction by A1,Lm57,NAT_6:2;
    end;
    hence thesis by A2;
  end;
