
theorem BBB:
for R being Ring
for q being Element of the carrier of Polynom-Ring R
for p being Polynomial of R
for n,j being Nat st p = n * q holds p.j = n * (q.j)
proof
let R be Ring, q be Element of the carrier of Polynom-Ring R;
let p be Polynomial of R; let n,j be Nat;
assume AS: p = n * q;
defpred P[Nat] means
   for q being Element of the carrier of Polynom-Ring R
   for p being Polynomial of R for j being Nat
   st p = ($1) * q holds p.j = ($1) * (q.j);
IA: P[0]
    proof
    now let q be Element of the carrier of Polynom-Ring R;
        let p be Polynomial of R; let j be Nat;
      assume p = 0 * q;
      then p = 0.(Polynom-Ring R) by BINOM:12
            .= 0_.(R) by POLYNOM3:def 10;
      hence p.j
         = 0.R by ORDINAL1:def 12,FUNCOP_1:7
        .= 0 * (q.j) by BINOM:12;
      end;
    hence thesis;
    end;
IS: now let k be Nat;
    assume IV: P[k];
    now let q be Element of the carrier of Polynom-Ring R;
        let p be Polynomial of R; let j be Nat;
      assume AS: p = (k+1) * q;
      reconsider p1 = k * q, r = q as Polynomial of R by POLYNOM3:def 10;
      p = k * q + 1 * q by AS,BINOM:15
        .= k * q + q by BINOM:13
        .= p1 + r by POLYNOM3:def 10;
      hence p.j
          = (p1.j) + (r.j) by NORMSP_1:def 2
         .= k * (q.j) + (q.j) by IV
         .= k * (q.j) + 1 * (q.j) by BINOM:13
         .= (k+1) * (q.j) by BINOM:15;
      end;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis by AS;
end;
