theorem Th4:
  for m, n being Nat holds
  card {k where k is Nat: m<=k & k<=m+n} = n+1
proof
  deffunc F(Nat) = $1;
  let m, n be Nat;
  defpred P[Nat] means m<=$1;
  set F = {F(k) where k is Nat: k<=m+n & P[k]};
  F is finite from FINSEQ_1:sch 6;
  then reconsider F as finite set;
  F = {F(k) where k is Nat: m<=k & k<=m+n}
  proof
    thus F c= {F(k) where k is Nat: m<=k & k<=m+n}
    proof
      let x be object;
      assume x in F;
      then ex k1 being Nat st x = F(k1) & k1 <= m+n & m <= k1;
      then ex k1 being Nat st x = F(k1) & m <= k1 & k1 <= m+n;
      hence thesis;
    end;
    let x be object;
    assume x in {F(k) where k is Nat: m<=k & k<=m+n};
    then ex k1 being Nat st x = F(k1) & m <= k1 & k1 <= m+n;
    hence thesis;
  end;
  then reconsider G = {F(k) where k is Nat: m<=k & k<=m+n} as finite set;
  card G = n+1 by Lm1;
  hence thesis;
end;
