
theorem XYZaa:
for F being Field
for m being Ordinal st m in card(nonConstantPolys F)
for p being non constant Polynomial of F
holds (Lt Poly(m,p)).m = deg p &
      for o being Ordinal st o <> m holds (Lt Poly(m,p)).o = 0
proof
let F be Field, m be Ordinal;
assume AS: m in card(nonConstantPolys F);
let p be non constant Polynomial of F;
set n = card(nonConstantPolys F), q = Poly(m,p);
for o being object st o in {m} holds o in n by AS,TARSKI:def 1;
then reconsider S = {m} as finite Subset of n by TARSKI:def 3;
reconsider degp = deg p as non zero Element of NAT by RATFUNC1:def 2;
H: m in {m} by TARSKI:def 1;
K: deg p = len p - 1 by HURWITZ:def 2; then
I: deg p = len p -' 1 by XREAL_0:def 2;
set b = (S,degp)-bag;
D: support b = S by UPROOTS:8;
E: b in Support q
   proof
   F: q.b = p.(b.m) by D,defPg .= p.degp by H,UPROOTS:7;
   p.degp = LC p by I,RATFUNC1:def 6; then
   p.degp <> 0.F;
   hence thesis by F,POLYNOM1:def 4;
   end;
for b1 being bag of n st b1 in Support q holds b1 <=' b
   proof
   let b1 be bag of n;
   assume b1 in Support q;
   then E1: q.b1 <> 0.F by POLYNOM1:def 4;
   per cases;
   suppose support b1 = {}; then
     b1 = EmptyBag n by PRE_POLY:81;
     hence thesis by PRE_POLY:60;
     end;
   suppose E0: support b1 = {m};
     per cases;
     suppose b1 = b;
       hence thesis;
       end;
     suppose EE: b1 <> b;
     E4: q.b1 = p.(b1.m) by E0,defPg;
     E3: b.m = len p - 1 by K,H,UPROOTS:7;
     E7: now assume b1.m >= b.m; then
        per cases by XXREAL_0:1;
        suppose E5: b1.m = b.m;
          now let i be object;
            assume i in n;
            per cases;
            suppose i = m;
              hence b1.i = b.i by E5;
              end;
            suppose i <> m;
              then E6: not i in {m} by TARSKI:def 1;
              hence b.i = 0 by UPROOTS:6 .= b1.i by E6,E0,PRE_POLY:def 7;
              end;
            end;
          hence contradiction by EE,PBOOLE:def 10;
          end;
        suppose b1.m > b.m;
          then b1.m >= (len p - 1) + 1 by E3,INT_1:7;
          hence contradiction by E4,E1,ALGSEQ_1:8;
          end;
        end;
     ex k being Ordinal st b1.k < b.k &
        for l being Ordinal st l in k holds b1.l = b.l
       proof
       take k = m;
       thus b1.k < b.k by E7;
       now let l be Ordinal;
         assume l in k;
         then l <> k;
         then E8: not l in {m} by TARSKI:def 1;
         hence b.l = 0 by UPROOTS:6 .= b1.l by E8,E0,PRE_POLY:def 7;
         end;
       hence thesis;
       end;
     hence thesis by PRE_POLY:def 9,PRE_POLY:def 10;
     end;
     end;
   suppose support b1 <> {} & support b1 <> {m};
     hence thesis by E1,defPg;
     end;
   end; then
F: b = Lt q by E,LT1;
hence (Lt Poly(m,p)).m = deg p by H,UPROOTS:7;
now let o be Ordinal;
  assume o <> m;
  then not o in {m} by TARSKI:def 1;
  hence (Lt Poly(m,p)).o = 0 by F,UPROOTS:6;
  end;
hence thesis;
end;
