Termination w.r.t. Q of the following Term Rewriting System could be proven:

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
  ↳ 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)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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)))

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

↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APPEND2(add2(N, X), Y) -> APPEND2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APPEND2(x1, x2)) = x1   
POL(add2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LT2(s1(X), s1(Y)) -> LT2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LT2(x1, x2)) = x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SPLIT2(N, add2(M, Y)) -> SPLIT2(N, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SPLIT2(x1, x2)) = x2   
POL(add2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


QSORT1(add2(N, X)) -> F_33(split2(N, X), N, X)
The remaining pairs can at least be oriented weakly.

F_33(pair2(Y, Z), N, X) -> QSORT1(Y)
F_33(pair2(Y, Z), N, X) -> QSORT1(Z)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F_33(x1, x2, x3)) = x1   
POL(QSORT1(x1)) = x1   
POL(add2(x1, x2)) = 1 + x2   
POL(f_14(x1, x2, x3, x4)) = 1 + x1   
POL(f_26(x1, x2, x3, x4, x5, x6)) = 1 + x5 + x6   
POL(false) = 0   
POL(lt2(x1, x2)) = 0   
POL(nil) = 0   
POL(pair2(x1, x2)) = x1 + x2   
POL(s1(x1)) = 0   
POL(split2(x1, x2)) = x2   
POL(true) = 0   

The following usable rules [14] were oriented:

f_26(false, N, M, Y, X, Z) -> pair2(add2(M, X), Z)
f_14(pair2(X, Z), N, M, Y) -> f_26(lt2(N, M), N, M, Y, X, Z)
split2(N, nil) -> pair2(nil, nil)
f_26(true, N, M, Y, X, Z) -> pair2(X, add2(M, Z))
split2(N, add2(M, Y)) -> f_14(split2(N, Y), N, M, Y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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) -> 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.