Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)


Q DP problem:
The TRS P consists of the following rules:

F_33(pair2(Y, Z), N, X) -> QSORT1(Y)
F_33(pair2(Y, Z), N, X) -> APPEND2(qsort1(Y), add2(X, qsort1(Z)))
F_14(pair2(X, Z), N, M, Y) -> F_26(lt2(N, M), N, M, Y, X, Z)
F_14(pair2(X, Z), N, M, Y) -> LT2(N, M)
QSORT1(add2(N, X)) -> F_33(split2(N, X), N, X)
LT2(s1(X), s1(Y)) -> LT2(X, Y)
SPLIT2(N, add2(M, Y)) -> F_14(split2(N, Y), N, M, Y)
APPEND2(add2(N, X), Y) -> APPEND2(X, Y)
F_33(pair2(Y, Z), N, X) -> QSORT1(Z)
QSORT1(add2(N, X)) -> SPLIT2(N, X)
SPLIT2(N, add2(M, Y)) -> SPLIT2(N, Y)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F_33(pair2(Y, Z), N, X) -> QSORT1(Y)
F_33(pair2(Y, Z), N, X) -> APPEND2(qsort1(Y), add2(X, qsort1(Z)))
F_14(pair2(X, Z), N, M, Y) -> F_26(lt2(N, M), N, M, Y, X, Z)
F_14(pair2(X, Z), N, M, Y) -> LT2(N, M)
QSORT1(add2(N, X)) -> F_33(split2(N, X), N, X)
LT2(s1(X), s1(Y)) -> LT2(X, Y)
SPLIT2(N, add2(M, Y)) -> F_14(split2(N, Y), N, M, Y)
APPEND2(add2(N, X), Y) -> APPEND2(X, Y)
F_33(pair2(Y, Z), N, X) -> QSORT1(Z)
QSORT1(add2(N, X)) -> SPLIT2(N, X)
SPLIT2(N, add2(M, Y)) -> SPLIT2(N, Y)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 5 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APPEND2(add2(N, X), Y) -> APPEND2(X, Y)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APPEND2(add2(N, X), Y) -> APPEND2(X, Y)
Used argument filtering: APPEND2(x1, x2)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LT2(s1(X), s1(Y)) -> LT2(X, Y)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LT2(s1(X), s1(Y)) -> LT2(X, Y)
Used argument filtering: LT2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SPLIT2(N, add2(M, Y)) -> SPLIT2(N, Y)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SPLIT2(N, add2(M, Y)) -> SPLIT2(N, Y)
Used argument filtering: SPLIT2(x1, x2)  =  x2
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F_33(pair2(Y, Z), N, X) -> QSORT1(Y)
QSORT1(add2(N, X)) -> F_33(split2(N, X), N, X)
F_33(pair2(Y, Z), N, X) -> QSORT1(Z)

The TRS R consists of the following rules:

lt2(0, s1(X)) -> true
lt2(s1(X), 0) -> false
lt2(s1(X), s1(Y)) -> lt2(X, Y)
append2(nil, Y) -> Y
append2(add2(N, X), Y) -> add2(N, append2(X, Y))
split2(N, nil) -> pair2(nil, nil)
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
qsort1(nil) -> nil
qsort1(add2(N, X)) -> f_33(split2(N, X), N, X)
f_33(pair2(Y, Z), N, X) -> append2(qsort1(Y), add2(X, qsort1(Z)))

The set Q consists of the following terms:

lt2(0, s1(x0))
lt2(s1(x0), 0)
lt2(s1(x0), s1(x1))
append2(nil, x0)
append2(add2(x0, x1), x2)
split2(x0, nil)
split2(x0, add2(x1, x2))
f_14(pair2(x0, x1), x2, x3, x4)
f_26(true, x0, x1, x2, x3, x4)
f_26(false, x0, x1, x2, x3, x4)
qsort1(nil)
qsort1(add2(x0, x1))
f_33(pair2(x0, x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.