
theorem lemlowp0:
for R being Ring,
    p being Polynomial of R
holds min* {i where i is Nat : (p+0_.R).i <> 0.R}
                                   = min* {i where i is Nat : p.i <> 0.R}
proof
let R be Ring, p be Polynomial of R;
now let o be object;
  assume o in {i where i is Nat : p.i <> 0.R};
  then consider i being Nat such that H1: o = i & p.i <> 0.R;
  thus o in NAT by H1,ORDINAL1:def 12;
  end;
then reconsider cp = {i where i is Nat : p.i <> 0.R} as Subset of NAT
  by TARSKI:def 3;
now let o be object;
  assume o in {i where i is Nat : (p+0_.R).i <> 0.R};
  then consider i being Nat such that H1: o = i & (p+0_.R).i <> 0.R;
  thus o in NAT by H1,ORDINAL1:def 12;
  end;
then reconsider cpq = {i where i is Nat : (p+0_.R).i <> 0.R} as Subset of NAT
  by TARSKI:def 3;
per cases;
suppose cp is non empty;
  then min* cp in cp by NAT_1:def 1;
  then consider i being Nat such that H1: min* cp = i & p.i <> 0.R;
  (p+0_.R).i = p.i + (0_.R).i by NORMSP_1:def 2
            .= p.i + 0.R by ORDINAL1:def 12,FUNCOP_1:7
            .= p.i;
  then B: min* cp in cpq by H1;
  now let k be Nat;
    assume k in cpq;
    then consider i being Nat such that C1: k = i & (p+0_.R).i <> 0.R;
    now assume min* cp > k;
      then C2: not(k in cp) by NAT_1:def 1;
      (p+0_.R).k = p.k + (0_.R).k by NORMSP_1:def 2
            .= p.k + 0.R by ORDINAL1:def 12,FUNCOP_1:7
            .= 0.R by C2;
      hence contradiction by C1;
      end;
    hence min* cp <= k;
    end;
  hence min*{i where i is Nat : (p+0_.R).i <> 0.R}
      = min*{i where i is Nat : p.i <> 0.R} by B,NAT_1:def 1;
  end;
suppose A: cp is empty;
  then p = 0_.(R) by lemlp0;
  hence min*{i where i is Nat : (p+0_.R).i <> 0.R}
      = min*{i where i is Nat : p.i <> 0.R} by A,lemlp0,POLYNOM3:28;
  end;
end;
