
theorem Th11:
for F being Field,
    a being non zero Element of F,
    i being Nat, m being Ordinal st m in card(nonConstantPolys F)
holds Poly(m,anpoly(a,0)) *' Poly(m,anpoly(1.F,i)) = Poly(m,anpoly(a,i))
proof
let F be Field, a be non zero Element of F, i be Nat, m being Ordinal;
set n = card(nonConstantPolys F);
assume AS: m in n;
   anpoly(a,0) = a|F by FIELD_1:7; then
C: Poly(m,anpoly(a,0)) = a|(n,F) by AS,XYZbb; then
A: Poly(m,anpoly(a,0)) *' Poly(m,anpoly(1.F,i))
      = a * Poly(m,anpoly(1.F,i)) by POLYNOM7:27;
set q = a * Poly(m,anpoly(1.F,i));
per cases;
suppose I: i = 0; then
  anpoly(1.F,i) = (1.F)|F by FIELD_1:7; then
  Poly(m,anpoly(1.F,i)) = (1.F)|(n,F) by AS,XYZbb;
  hence Poly(m,anpoly(a,0)) *' Poly(m,anpoly(1.F,i))
      = a|(n,F) *' 1_(n,F) by C,POLYNOM7:20
     .= Poly(m,anpoly(a,i)) by I,C,POLYNOM1:29;
  end;
suppose I: i <> 0;
now let o be Element of Bags n;
  per cases;
  suppose support o = {};
    then B: o = EmptyBag n by PRE_POLY:81;
    hence Poly(m,anpoly(a,i)).o
        = anpoly(a,i).0 by defPg
       .= a * 0.F by I,POLYDIFF:25
       .= a * (anpoly(1.F,i).0) by I,POLYDIFF:25
       .= a * (Poly(m,anpoly(1.F,i))).o by B,defPg
       .= q.o by POLYNOM7:def 9;
    end;
  suppose I1: support o = {m};
    per cases;
    suppose I2: o.m = i;
      thus Poly(m,anpoly(a,i)).o
         = anpoly(a,i).(o.m) by I1,defPg
        .= a * 1.F by I2,POLYDIFF:24
        .= a * anpoly(1.F,i).(o.m) by I2,POLYDIFF:24
        .= a * Poly(m,anpoly(1.F,i)).o by I1,defPg
        .= q.o by POLYNOM7:def 9;
      end;
    suppose I2: o.m <> i;
      thus Poly(m,anpoly(a,i)).o
         = anpoly(a,i).(o.m) by I1,defPg
        .= a * 0.F by I2,POLYDIFF:25
        .= a * anpoly(1.F,i).(o.m) by I2,POLYDIFF:25
        .= a * Poly(m,anpoly(1.F,i)).o by I1,defPg
        .= q.o by POLYNOM7:def 9;
      end;
    end;
  suppose I2: support o <> {} & support o <> {m};
    hence Poly(m,anpoly(a,i)).o
        = a * 0.F by defPg
       .= a * Poly(m,anpoly(1.F,i)).o by I2,defPg
       .= q.o by POLYNOM7:def 9;
    end;
  end;
hence thesis by A;
end;
end;
