
theorem Th12:
  for m,n being non zero Nat st m >= n holds Lucas m >= Lucas n
  proof
    let m,n be non zero Nat;
    assume A1: m >= n;
    per cases by A1,XXREAL_0:1;
    suppose m = n;
      hence thesis;
    end;
    suppose A2: m > n;
    then consider k being Nat such that A3: m = n + k by NAT_1:10;
A4: for k, n being non zero Nat holds Lucas (n+k) >= Lucas (n)
    proof
      defpred P[Nat] means
      for n being non zero Nat holds Lucas (n+$1) >= Lucas (n);
A5:   P[1]
      proof
        let n be non zero Nat;
        n - 0 is Element of NAT & n+1 is Element of NAT by ORDINAL1:def 12;
        hence thesis by FIB_NUM3:18;
      end;
A6:   for k being non zero Nat st P[k] holds P[k + 1]
      proof
        let k be non zero Nat;
        assume A7: P[k];
        for n being non zero Nat holds Lucas (n + (k + 1)) >= Lucas n
        proof
          let n be non zero Nat;
          reconsider p = n + k as Element of NAT by ORDINAL1:def 12;
          p is non zero Element of NAT; then
          Lucas (n + k) >= Lucas n & Lucas (n + k + 1) >= Lucas (n + k)
            by A7,FIB_NUM3:18;
          hence thesis by XXREAL_0:2;
        end;
        hence thesis;
      end;
      for k being non zero Nat holds P[k] from NAT_1:sch 10(A5,A6);
      hence thesis;
    end;
    k is non zero Nat & n is non zero Nat by A3,A2;
    hence thesis by A3,A4;
    end;
  end;
