
theorem T4:
for R being Ring
for a,b being Element of R holds (a|R) *' (b|R) = (a * b)|R
proof
let R be Ring, a,b being Element of R;
set p = a|R, q = b|R;
now let x be object;
  assume x in NAT;
  then reconsider i = x as Element of NAT;
  consider F being FinSequence of the carrier of R such that
  H: len F = i+1 & (p*'q).i = Sum F &
     for k being Element of NAT st k in dom F
     holds F.k = p.(k-'1) * q.(i+1-'k) by POLYNOM3:def 9;
  per cases;
  suppose A: i = 0;
    then 1 in Seg(len F) by H,FINSEQ_1:1;
    then 1 in dom F by FINSEQ_1:def 3;
    then F.1 = p.(0+1-'1) * q.(i+1-'1) by H
            .= p.0 * q.(i+1-'1) by NAT_D:34
            .= p.0 * q.0 by NAT_D:34,A
            .= a * q.0 by Th28
            .= a * b by Th28;
    then F = <*a*b*> by FINSEQ_1:40,A,H;
    hence (p*'q).x = a * b by H,RLVECT_1:44 .= (a * b)|R.x by A,Th28;
    end;
  suppose A: i > 0;
    now let j be Element of NAT;
      assume B: j in dom F;
      then j in Seg(len F) by FINSEQ_1:def 3;
      then C: 1 <= j & j <= i+1 by H,FINSEQ_1:1;
      p.(j-'1) = 0.R or q.(i+1-'j) = 0.R
        proof
        assume p.(j-'1) <> 0.R;
        then j <= 1 by NAT_D:36,Th28;
        then j = 1 by C,XXREAL_0:1;
        then i + 1 -'j = i by NAT_D:34;
        hence q.(i+1-'j) = 0.R by A,Th28;
        end;
      hence 0.R = p.(j-'1) * q.(i+1-'j) .= F.j by B,H;
      end;
    hence (p*'q).x = 0.R by H,POLYNOM3:1 .= (a * b)|R.x by A,Th28;
    end;
  end;
hence thesis by FUNCT_2:12;
end;
