Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔)
        ↳24 TRUE

(0) Obligation:

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

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

LT(s(X), s(Y)) → LT(X, Y)
APPEND(add(N, X), Y) → APPEND(X, Y)
SPLIT(N, add(M, Y)) → F_1(split(N, Y), N, M, Y)
SPLIT(N, add(M, Y)) → SPLIT(N, Y)
F_1(pair(X, Z), N, M, Y) → F_2(lt(N, M), N, M, Y, X, Z)
F_1(pair(X, Z), N, M, Y) → LT(N, M)
QSORT(add(N, X)) → F_3(split(N, X), N, X)
QSORT(add(N, X)) → SPLIT(N, X)
F_3(pair(Y, Z), N, X) → APPEND(qsort(Y), add(X, qsort(Z)))
F_3(pair(Y, Z), N, X) → QSORT(Y)
F_3(pair(Y, Z), N, X) → QSORT(Z)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APPEND(add(N, X), Y) → APPEND(X, Y)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APPEND(add(N, X), Y) → APPEND(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APPEND(x1, x2)) = x1   
POL(add(x1, x2)) = 1 + x2   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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

LT(s(X), s(Y)) → LT(X, Y)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


LT(s(X), s(Y)) → LT(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(LT(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

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

SPLIT(N, add(M, Y)) → SPLIT(N, Y)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SPLIT(N, add(M, Y)) → SPLIT(N, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(SPLIT(x1, x2)) = x2   
POL(add(x1, x2)) = 1 + x2   

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

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

QSORT(add(N, X)) → F_3(split(N, X), N, X)
F_3(pair(Y, Z), N, X) → QSORT(Y)
F_3(pair(Y, Z), N, X) → QSORT(Z)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QSORT(add(N, X)) → F_3(split(N, X), N, X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F_3(x1, x2, x3)) = x1   
POL(QSORT(x1)) = x1   
POL(add(x1, x2)) = 1 + x2   
POL(f_1(x1, x2, x3, x4)) = 1 + x1   
POL(f_2(x1, x2, x3, x4, x5, x6)) = 1 + x5 + x6   
POL(false) = 0   
POL(lt(x1, x2)) = 0   
POL(nil) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(s(x1)) = 0   
POL(split(x1, x2)) = x2   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)

(22) Obligation:

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

F_3(pair(Y, Z), N, X) → QSORT(Y)
F_3(pair(Y, Z), N, X) → QSORT(Z)

The TRS R consists of the following rules:

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
append(add(N, X), Y) → add(N, append(X, Y))
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(24) TRUE