
theorem thX3:
for p being Prime
for n being non zero Nat
for F being p-characteristic Field holds (Deriv F).X^(p|^n,F) = -1_.(F)
proof
let p be Prime, n be non zero Nat, F be p-characteristic Field;
set q = X^(p|^n,F), Dq = (Deriv F).q;
now let m be Nat;
  per cases;
  suppose A: m = 0;
    hence Dq.m = (0+1) * q.(0+1) by RINGDER1:def 8
              .= 1 * (-1.F) by Lm10
              .= -(1.F) by BINOM:13
              .= -((1_.(F)).m) by A,POLYNOM3:30
               .= (-1_.(F)).m by BHSP_1:44;
    end;
  suppose A: m = 1;
    per cases;
    suppose B: p = 2 & n = 1; then
      C: Char F = 2 by RING_3:def 6;
      thus Dq.m = (1+1) * q.(1+1) by A,RINGDER1:def 8
               .= 2 * 1.F by B,Lm10
               .= 2 '*' 1.F by RING_3:def 2
               .= -0.F by C,REALALG2:24
               .= -((1_.(F)).m) by A,POLYNOM3:30
               .= (-1_.(F)).m by BHSP_1:44;
      end;
    suppose B: p <> 2 or n <> 1;
      C: 2 <> p|^n
         proof
         p > 1 by INT_2:def 4; then
         E: p >= 1 + 1 by INT_1:7;
         F: n >= 0 + 1 by INT_1:7;
         per cases by B;
         suppose p <> 2;
           then p > 2 by E,XXREAL_0:1;
           then p >= 2 + 1 by INT_1:7;
           then H: p|^n >= 3|^n by thX3a;
           3|^n >= 3|^1 by F,NAT_6:1;
           hence thesis by H,XXREAL_0:2;
           end;
         suppose n <> 1;
           then H: n is non trivial by NAT_2:def 1;
           I: p|^n >= 2|^n by E,thX3a;
           2|^n >= 2|^2 by H,NAT_2:29,NAT_6:1;
           then p|^n >= 2|^(1+1) by I,XXREAL_0:2;
           then p|^n >= 2|^1 * 2 by NEWTON:6;
           hence thesis;
           end;
         end;
      thus Dq.m = (1+1) * q.(1+1) by A,RINGDER1:def 8
               .= 2 * 0.F by C,Lm11
               .= -0.F
               .= -((1_.(F)).m) by A,POLYNOM3:30
               .= (-1_.(F)).m by BHSP_1:44;
      end;
    end;
  suppose A: m <> 0 & m <> 1 & m = p|^n - 1;
    reconsider n1 = n - 1 as Nat;
    B: p|^(n1+1) = p|^n1 * p by NEWTON:6;
    thus Dq.m = p|^n * q.(p|^n-1+1) by A,RINGDER1:def 8
             .= -0.F by B,FIELD_15:37
             .= -((1_.(F)).m) by A,POLYNOM3:30
             .= (-1_.(F)).m by BHSP_1:44;
    end;
  suppose A: m <> 0 & m <> 1 & m <> p|^n - 1; then
    B: m + 1 <> 1 & m + 1 <> p|^n;
    thus Dq.m = (m+1) * q.(m+1) by RINGDER1:def 8
             .= (m+1) * 0.F by B,Lm11
             .= -0.F
             .= -((1_.(F)).m) by A,POLYNOM3:30
             .= (-1_.(F)).m by BHSP_1:44;
    end;
  end;
hence thesis;
end;
