
theorem XYZbb:
for F being Field
for m being Ordinal st m in card(nonConstantPolys F)
for p being Polynomial of F
for a being Element of F
holds Poly(m,p) = a|(card(nonConstantPolys F),F) iff p = a|F
proof
let F be Field, m be Ordinal;
assume K: m in card(nonConstantPolys F);
let p be Polynomial of F; let a be Element of F;
set n = card(nonConstantPolys F);
A: now assume AS: p = a|F;
   set q = a|(n,F);
   now let o be object;
     assume o in Bags n;
     then reconsider b = o as bag of n;
     per cases;
     suppose support b = {};
       then for i being object st i in n holds b.i = {} by PRE_POLY:def 7;
       then D: b = EmptyBag n by PBOOLE:6;
       hence Poly(m,p).o = p.0 by defPg
                        .= a by AS,Th28
                        .= q.o by D,POLYNOM7:18;
       end;
     suppose D: support b = {m};
       m in {m} by TARSKI:def 1; then
       E: b.m <> 0 by D,PRE_POLY:def 7;
       F: b <> EmptyBag n by D;
       thus Poly(m,p).o = p.(b.m) by D,defPg
                       .= 0.F by AS,E,Th28
                       .= q.o by F,POLYNOM7:18;
       end;
     suppose E: support b <> {} & support b <> {m};
       then D: b <> EmptyBag n;
       thus Poly(m,p).o = 0.F by E,defPg
                       .= q.o by D,POLYNOM7:18;
       end;
     end;
   hence Poly(m,p) = a|(n,F);
   end;
now assume AS: Poly(m,p) = a|(n,F);
   now let o be object;
     assume o in NAT;
     then reconsider i = o as Element of NAT;
     per cases;
     suppose B: i = 0;
       then p.i = Poly(m,p).(EmptyBag n) by defPg
               .= a by AS,POLYNOM7:18
               .= (a|F).i by B,Th28;
       hence p.o = (a|F).o;
       end;
     suppose C: i <> 0;
       for o being object st o in {m} holds o in n by K,TARSKI:def 1;
       then reconsider S = {m} as finite Subset of n by TARSKI:def 3;
       set b = (S,i)-bag;
       B: support b = S by C,UPROOTS:8; then
       D: b <> EmptyBag n;
       m in {m} by TARSKI:def 1;
       then p.i = p.(b.m) by UPROOTS:7
               .= Poly(m,p).b by B,defPg
               .= 0.F by D,AS,POLYNOM7:18
               .= (a|F).i by C,Th28;
       hence p.o = (a|F).o;
       end;
     end;
   hence p = a|F;
   end;
hence thesis by A;
end;
