
theorem lemlowp3a:
for R being non degenerated Ring,
    p,q being non zero Polynomial of R
st p *' q <> 0_.(R)
holds min* {i where i is Nat : (p*'q).i <> 0.R} >=
                   min* {i where i is Nat : p.i <> 0.R} +
                   min* {i where i is Nat : q.i <> 0.R}
proof
let R be non degenerated Ring, p,q be non zero Polynomial of R;
reconsider cp = {i where i is Nat : p.i <> 0.R},
           cq = {i where i is Nat : q.i <> 0.R} as
                             non empty Subset of NAT by lemlp1;
assume p *' q <> 0_.(R);
then p *' q is non zero;
then reconsider cpq = {i where i is Nat : (p*'q).i <> 0.R} as
                      non empty Subset of NAT by lemlp1;
set m = min* cp + min* cq;
A: min* cpq in cpq by NAT_1:def 1;
now let z1 be Nat;
  assume z1 in cpq;
  then consider w being Nat such that H1: w = z1 & (p*'q).w <> 0.R;
  reconsider z = z1 as Element of NAT by ORDINAL1:def 12;
  consider r1 being FinSequence of the carrier of R such that
  R: len r1 = z+1 & (p*'q).z = Sum r1 &
     for k be Element of NAT st k in dom r1 holds r1.k = p.(k-'1) * q.(z+1-'k)
     by POLYNOM3:def 9;
  1 <= z + 1 by NAT_1:11;
  then A1: 1 in dom r1 by FINSEQ_3:25,R;
  now assume AA1: z < m;
    F: for k being Nat st k in dom r1 holds p.(k-'1) = 0.R or q.(z+1-'k) = 0.R
       proof
       let k be Nat;
       assume k in dom r1;
       then EF: 1 <= k & k <= len r1 by FINSEQ_3:25;
       per cases;
       suppose AA0: k-'1 >= min* cp;
         AA2: k -' 1 = k - 1 by EF,XREAL_0:def 2;
         z + 1 - k in NAT by INT_1:5,R,EF;
         then z + 1 -' k = z + 1 - k by XREAL_0:def 2;
         then (k -' 1) + (z + 1 -' k) < min* cp + min* cq by AA1,AA2;
         then S: z+1-'k < min* cq by AA0,XREAL_1:7;
         now assume q.(z+1-'k) <> 0.R;
           then z+1-'k in cq;
           hence contradiction by S,NAT_1:def 1;
           end;
         hence thesis;
         end;
       suppose S: k-'1 < min* cp;
         now assume p.(k-'1) <> 0.R;
           then k-'1 in cp;
           hence contradiction by S,NAT_1:def 1;
           end;
         hence thesis;
         end;
       end;
    now let k be Element of NAT;
      assume E: k in dom r1 & k <> 1;
      per cases by F,E;
      suppose p.(k-'1) = 0.R;
        then r1.k = 0.R * q.(z+1-'k) by E,R;
        hence r1/.k = 0.R by E,PARTFUN1:def 6;
        end;
      suppose q.(z+1-'k) = 0.R;
        then r1.k = p.(k-'1) * 0.R by E,R;
        hence r1/.k = 0.R by E,PARTFUN1:def 6;
        end;
      end;
    then T: Sum r1 = r1/.1 by A1,POLYNOM2:3
                  .= r1.1 by A1,PARTFUN1:def 6
                  .= p.(1-'1) * q.(z+1-'1) by A1,R;
    now per cases by A1,F;
      suppose p.(1-'1) = 0.R;
        hence Sum r1 = 0.R by T;
      end;
      suppose q.(z+1-'1) = 0.R;
        hence Sum r1 = 0.R by T;
      end;
      end;
    hence contradiction by H1,R;
    end;
  hence z1 >= m;
  end;
hence thesis by A;
end;
