
theorem lem5:
for f being ascending Field-yielding sequence
for i being Element of NAT
holds 1.(SeqField f) = 1.(f.i) & 0.(SeqField f) = 0.(f.i)
proof
let f be ascending Field-yielding sequence, i be Element of NAT;
defpred P[Nat] means
  ex k being Element of NAT st k = $1 & 1.(f.k) = 1.(f.0) & 0.(f.k) = 0.(f.0);
IA: P[0];
IS: now let k be Nat;
    assume P[k]; then
    consider n being Element of NAT such that
    IV: k = n & 1.(f.n) = 1.(f.0) & 0.(f.n) = 0.(f.0);
    f.(n+1) is FieldExtension of f.n by defasc; then
    f.n is Subfield of f.(n+1) by FIELD_4:7; then
    1.(f.n) = 1.(f.(n+1)) & 0.(f.n) = 0.(f.(n+1)) by EC_PF_1:def 1;
    hence P[k+1] by IV;
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
then P[i] & 1.(SeqField f) = 1.(f.0) & 0.(SeqField f) = 0.(f.0) by dsf;
hence 1.(SeqField f) = 1.(f.i) & 0.(SeqField f) = 0.(f.i);
end;
