
theorem div3a:
for F being Field,
    p,r being Polynomial of F
for q being non zero Polynomial of F
holds (p *' r) mod q = ((p mod q) *' (r mod q)) mod q
proof
let F be Field, p,r be Polynomial of F;
let q be non zero Polynomial of F;
H1: (((p div q) *' q) *' ((r div q) *' q) + ((p div q) *' q) *' (r mod q))
  = ((((p div q) *' q) *' (r div q))) *' q + ((p div q) *' q) *' (r mod q)
     by POLYNOM3:33
 .= q *' ((((p div q) *' q) *' (r div q))) + q *' ((p div q) *' (r mod q))
    by POLYNOM3:33
 .= q *' (((((p div q) *' q) *' (r div q))) + (p div q) *' (r mod q))
    by POLYNOM3:31;
H2: (p mod q) *' ((r div q) *' q) = q *' ((p mod q) *' (r div q))
    by POLYNOM3:33;
p *' r = ((p div q) *' q + (p mod q)) *' r by div5
  .= ((p div q) *' q + (p mod q)) *' ((r div q) *' q + (r mod q)) by div5
  .= ((p div q) *' q) *' ((r div q) *' q + (r mod q)) +
     (p mod q) *' ((r div q) *' q + (r mod q)) by POLYNOM3:31
  .= (((p div q) *' q) *' ((r div q) *' q) +
      ((p div q) *' q) *' (r mod q)) +
     (p mod q) *' ((r div q) *' q + (r mod q)) by POLYNOM3:31
  .= (((p div q) *' q) *' ((r div q) *' q) +
      ((p div q) *' q) *' (r mod q)) +
     ((p mod q) *' ((r div q) *' q) +
       (p mod q) *' (r mod q)) by POLYNOM3:31;
hence (p *' r) mod q
   = ((((p div q) *' q) *' ((r div q) *' q) +
      ((p div q) *' q) *' (r mod q)) mod q) +
     (((p mod q) *' ((r div q) *' q) +
       (p mod q) *' (r mod q)) mod q) by div4
  .= 0_.(F) +
     (((p mod q) *' ((r div q) *' q) +
       (p mod q) *' (r mod q)) mod q) by div2,H1,T2
  .= (((p mod q) *' ((r div q) *' q) +
       (p mod q) *' (r mod q)) mod q) by POLYNOM3:28
  .= (((p mod q) *' ((r div q) *' q)) mod q) +
     (((p mod q) *' (r mod q)) mod q) by div4
  .= 0_.(F) +
     (((p mod q) *' (r mod q)) mod q) by div2,H2,T2
  .= ((p mod q) *' (r mod q)) mod q by POLYNOM3:28;
end;
