
theorem Th5:
  for n be Element of NAT for p,q,r be Element of n-tuples_on NAT
  holds (p < q & q < r implies p < r) & ( p < q & q <= r or p <= q & q < r or p
  <= q & q <= r implies p <= r)
proof
  let n be Element of NAT;
  let p,q,r be Element of n-tuples_on NAT;
  thus
A1: p < q & q < r implies p < r
  proof
    assume that
A2: p < q and
A3: q < r;
    consider i be Element of NAT such that
A4: i in Seg n and
A5: p.i < q.i and
A6: for k be Nat st 1 <= k & k < i holds p.k = q.k by A2;
    consider j be Element of NAT such that
A7: j in Seg n and
A8: q.j < r.j and
A9: for k be Nat st 1 <= k & k < j holds q.k = r.k by A3;
    reconsider t = min(i,j) as Element of NAT;
    take t;
    thus t in Seg n by A4,A7,XXREAL_0:15;
    now
      per cases by XXREAL_0:1;
      suppose
A10:    i < j;
A11:    i >= 1 by A4,FINSEQ_1:1;
        t = i by A10,XXREAL_0:def 9;
        hence p.t < r.t by A5,A9,A10,A11;
      end;
      suppose
A12:    i > j;
A13:    j >= 1 by A7,FINSEQ_1:1;
        t = j by A12,XXREAL_0:def 9;
        hence p.t < r.t by A6,A8,A12,A13;
      end;
      suppose
        i = j;
        hence p.t < r.t by A5,A8,XXREAL_0:2;
      end;
    end;
    hence p.t < r.t;
    let k be Nat;
    assume that
A14: 1 <= k and
A15: k < t;
    t <= j by XXREAL_0:17;
    then
A16: k < j by A15,XXREAL_0:2;
    t <= i by XXREAL_0:17;
    then k < i by A15,XXREAL_0:2;
    hence p.k = q.k by A6,A14
      .= r.k by A9,A14,A16;
  end;
  assume
A17: p < q & q <= r or p <= q & q < r or p <= q & q <= r;
  per cases by A17;
  suppose
A18: p < q & q <= r;
    thus thesis by A1,A18;
  end;
  suppose
A19: p <= q & q < r;
    thus thesis by A1,A19;
  end;
  suppose
    p <= q & q <= r;
    hence thesis by A1;
  end;
end;
