reserve X for set;

theorem Th7:
  for S being non empty Subset-Family of X, N, G, F being sequence of S holds
  G.0 = N.0 & (for n being Nat holds G.(n+1) = N.(n+1)
\/ G.n) & F.0 = N.0 & (for n being Nat holds F.(n+1) = N.(n+1) \ G.n
  ) implies for n,m being Nat st n <= m holds F.n c= G.m
proof
  let S be non empty Subset-Family of X, N, G, F be sequence of S;
  assume that
A1: G.0 = N.0 and
A2: for n being Nat holds G.(n+1) = N.(n+1) \/ G.n and
A3: F.0 = N.0 and
A4: for n being Nat holds F.(n+1) = N.(n+1) \ G.n;
  let n,m be Nat;
A5: for n being Nat holds F.n c= G.n
  proof
    defpred P[Nat] means F.$1 c= G.$1;
A6: for n being Nat st P[n] holds P[n+1]
    proof
      let n be Nat;
      assume F.n c= G.n;
      G.(n+1) = N.(n+1) \/ G.n by A2;
      then
A7:   N.(n+1) c= G.(n+1) by XBOOLE_1:7;
      F.(n+1) = N.(n+1) \ G.n by A4;
      hence thesis by A7,XBOOLE_1:1;
    end;
A8: P[0] by A1,A3;
    thus for n being Nat holds P[n] from NAT_1:sch 2(A8,A6);
  end;
A9: n < m implies F.n c= G.m
  proof
    assume n < m;
    then
A10: G.n c= G.m by A1,A2,Th6;
    F.n c= G.n by A5;
    hence thesis by A10;
  end;
  assume n <= m;
  then n = m or n < m by XXREAL_0:1;
  hence thesis by A5,A9;
end;
