
theorem Th16:
  for n, k being Element of NAT holds PFCrt (n+1,k) = PFCrt (n,k)
  \/ {[2*n+3,k]}
proof
  let n,k be Element of NAT;
A1: 2*(n+1)+1 = 2*n+3;
  thus PFCrt (n+1,k) c= PFCrt (n,k) \/ {[2*n+3,k]}
  proof
    let x be object;
    assume x in PFCrt (n+1,k);
    then consider m being odd Element of NAT such that
A2: m <= 2*(n+1) + 1 and
A3: [m,k] = x by Def3;
    per cases by A2,NAT_1:9;
    suppose
      m <= 2*(n+1);
      then m <= 2*n + 1 or m = 2*n + 1 + 1 by NAT_1:8;
      then x in PFCrt (n,k) by A3,Def3;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose
      m = 2*(n+1) + 1;
      then x in {[2*n+3,k]} by A3,TARSKI:def 1;
      hence thesis by XBOOLE_0:def 3;
    end;
  end;
  let x be object;
  assume
A4: x in PFCrt (n,k) \/ {[2*n+3,k]};
  per cases by A4,XBOOLE_0:def 3;
  suppose
    x in PFCrt (n,k);
    then consider m being odd Element of NAT such that
A5: m <= 2*n + 1 and
A6: [m,k] = x by Def3;
    2*n+1 <= 2*(n+1)+1 by Lm4;
    then m <= 2*(n+1) + 1 by A5,XXREAL_0:2;
    hence thesis by A6,Def3;
  end;
  suppose
    x in {[2*n+3,k]};
    then x = [2*n+3,k] by TARSKI:def 1;
    hence thesis by A1,Def3;
  end;
end;
