
theorem Th13:
for F being Field,
    i being Nat
for m being Ordinal st m in card(nonConstantPolys F) holds
(power Polynom-Ring(card(nonConstantPolys F),F)).(Poly(m,anpoly(1.F,1)),i)
    = Poly(m,anpoly(1.F,i))
proof
let F be Field, i be Nat, m be Ordinal;
set n = card(nonConstantPolys F);
set f = power Polynom-Ring(card(nonConstantPolys F),F);
assume AS: m in n;
defpred P[Nat] means
  f.(Poly(m,anpoly(1.F,1)),$1) = Poly(m,anpoly(1.F,$1));
reconsider p = Poly(m,anpoly(1.F,1)) as Element of Polynom-Ring(n,F)
    by POLYNOM1:def 11;
IA: P[0]
    proof
    anpoly(1.F,0) = (1.F)|F by FIELD_1:7; then
    Poly(m,anpoly(1.F,0))
            = (1.F)|(n,F) by AS,XYZbb
           .= 1_(n,F) by POLYNOM7:20
           .= 1_(Polynom-Ring(n,F)) by POLYNOM1:def 11
           .= f.(p,0) by GROUP_1:def 7;
    hence thesis;
    end;
IS: now let k be Nat;
    assume IV: P[k];
    H: k is Element of NAT by ORDINAL1:def 12;
    reconsider q = Poly(m,anpoly(1.F,k)) as Element of Polynom-Ring(n,F)
         by POLYNOM1:def 11;
    f.(p,k+1)
       = f.(p,k) * p by GROUP_1:def 7
      .= Poly(m,anpoly(1.F,k)) *' Poly(m,anpoly(1.F,1)) by IV,POLYNOM1:def 11
      .= Poly(m,anpoly(1.F,k+1)) by H,AS,Th13a;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis;
end;
