
theorem lemlowp1a1:
for R being non degenerated Ring,
    p,q being non zero Polynomial of R
st min* {i where i is Nat : p.i <> 0.R} > min* {i where i is Nat : q.i <> 0.R}
holds min* {i where i is Nat : (p+q).i <> 0.R} =
                                          min* {i where i is Nat : q.i <> 0.R}
proof
let R be non degenerated Ring;
let p,q be non zero Polynomial of R;
assume AS1: min*{i where i is Nat : p.i<>0.R} >
            min*{i where i is Nat : q.i<>0.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;
now let o be object;
  assume o in {i where i is Nat : (p+q).i <> 0.R};
  then consider i being Nat such that H1: o = i & (p+q).i <> 0.R;
  thus o in NAT by H1,ORDINAL1:def 12;
  end;
then reconsider cpq = {i where i is Nat : (p+q).i <> 0.R} as Subset of NAT
  by TARSKI:def 3;
  not( (min* cq) in cp ) by AS1,NAT_1:def 1;
  then p.(min* cq) = 0.R;
  then A: (p+q).(min* cq) = 0.R + q.(min* cq) by NORMSP_1:def 2;
  min* cq in cq by NAT_1:def 1;
  then consider u being Nat such that H1: u = min* cq & q.u <> 0.R;
  B: min* cq in cpq by A,H1;
  then reconsider cpq as non empty Subset of NAT;
  now let k be Nat;
    assume k in cpq;
    then consider v being Nat such that H2: v = k & (p+q).v <> 0.R;
    now let j be Nat;
       assume D0: j < min* cq;
       D1: now assume q.j <> 0.R;
           then j in cq;
           hence contradiction by D0,NAT_1:def 1;
           end;
       now assume p.j <> 0.R;
           then j in cp;
           then j >= min* cp by NAT_1:def 1;
           hence contradiction by D0,AS1,XXREAL_0:2;
           end;
       hence (p+q).j = 0.R + q.j by NORMSP_1:def 2
                    .= 0.R by D1;
       end;
    hence min* cq <= k by H2;
    end;
  hence thesis by B,NAT_1:def 1;
end;
