theorem Th6:
 for k,n st n <= k holds n-th_FCEx(Al) c= k-th_FCEx(Al)
proof
  let k;
  defpred P[Nat] means $1 <= k implies ex j st j = k-$1 &
   j-th_FCEx(Al) c= k-th_FCEx(Al);
A1: P[0];
A2: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat such that
A3:  P[n];
    per cases;
    suppose
A4:   n+1 <= k;
      then consider j such that
A5:    j = k-n & j-th_FCEx(Al) c= k-th_FCEx(Al) by A3,NAT_1:13;
      set j2=k-(n+1);
      reconsider j2 as Element of NAT by A4,NAT_1:21;
      FCEx(j2-th_FCEx(Al)) = j-th_FCEx(Al) by A5,Th5;
      then j2-th_FCEx(Al) c= j-th_FCEx(Al) by QC_TRANS:def 1;
      hence thesis by A5,XBOOLE_1:1;
    end;
    suppose not n+1 <= k;
      hence thesis;
    end;
  end;
A6: for n being Nat holds P[n] from NAT_1:sch 2(A1,A2);
  let n such that
A7: n <= k;
  set n2 = k - n;
  reconsider n2 as Element of NAT by A7,NAT_1:21;
  k = n + n2;
  then consider n3 being Element of NAT such that
A8: n3 = k-n2 & n3-th_FCEx(Al) c= k-th_FCEx(Al) by A6,NAT_1:11;
  thus thesis by A8;
end;
