
theorem Lm0a:
for p being Prime
for R being p-characteristic commutative Ring
for a being non zero Element of R
for n being non zero Nat st n < p holds n * a <> 0.R
proof
let p be Prime, R be p-characteristic commutative Ring;
let a be non zero Element of R, n be non zero Nat;
assume AS: n < p;
defpred P[Nat] means $1 <> 0 & ($1) * a = 0.R;
P[p] by Lm0; then
E: ex k being Nat st P[k];
consider u being Nat such that
D: P[u] & for v being Nat st P[v] holds u <= v from NAT_1:sch 5(E);
Z: P[p] by Lm0; then
Y: u <= p by D;
now assume G0: u < p;
  defpred Q[Nat] means ($1) * u <= p;
  IA: Q[1] by Z,D;
  IS: now let k be Nat;
      assume 1 <= k;
      assume Q[k];
      then reconsider v = p - k * u as Element of NAT by INT_1:5;
      G: now assume v = 0;
         then u divides p by NAT_D:def 3;
         then u = 1 by G0,INT_2:def 4;
         hence contradiction by D,BINOM:13;
         end;
      consider m being Nat such that
      G4: (v = m & v '*' a = m*a) or (v = -m & v '*' a = -(m*a))
          by RING_3:def 2;
      consider m1 being Nat such that
      G5: (u = m1 & u '*' a = m1*a) or (u = -m1 & u '*' a = -(m1*a))
          by RING_3:def 2;
      consider m2 being Nat such that
      G6: (p = m2 & p '*' a = m2*a) or (p = -m2 & p '*' a = -(m2*a))
          by RING_3:def 2;
      v * a = p '*' a - (k * u) '*' a by G4,G,RING_3:64
           .= p '*' a - k '*' (u '*' a) by RING_3:65
           .= p '*' a + -0.R by D,G5,Lm0x
           .= 0.R by G6,Lm0;
      then u + k * u <= p - k * u + k * u by G,D,XREAL_1:6;
      hence Q[k+1];
      end;
  I: for k being Nat st k >= 1 holds Q[k] from NAT_1:sch 8(IA,IS);
  G1: p >= 1 by INT_2:def 4;
  u = 1
     proof
     G2: u >= 0 + 1 by D,INT_1:7;
     now assume u > 1;
       then u * p > 1 * p by XREAL_1:68;
       hence contradiction by G1,I;
       end;
     hence thesis by G2,XXREAL_0:1;
     end;
  hence contradiction by D,BINOM:13;
  end;
then u = p by Y,XXREAL_0:1;
hence n * a <> 0.R by D,AS;
end;
