Term Rewriting System R:
[X, Y, N, M, Z]
lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LT(s(X), s(Y)) -> LT(X, Y)
APPEND(add(N, X), Y) -> APPEND(X, Y)
SPLIT(N, add(M, Y)) -> F1(split(N, Y), N, M, Y)
SPLIT(N, add(M, Y)) -> SPLIT(N, Y)
F1(pair(X, Z), N, M, Y) -> F2(lt(N, M), N, M, Y, X, Z)
F1(pair(X, Z), N, M, Y) -> LT(N, M)
QSORT(add(N, X)) -> F3(split(N, X), N, X)
F3(pair(Y, Z), N, X) -> APPEND(qsort(Y), add(X, qsort(Z)))
F3(pair(Y, Z), N, X) -> QSORT(Y)
F3(pair(Y, Z), N, X) -> QSORT(Z)

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

LT(s(X), s(Y)) -> LT(X, Y)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LT(s(X), s(Y)) -> LT(X, Y)
one new Dependency Pair is created:

LT(s(s(X'')), s(s(Y''))) -> LT(s(X''), s(Y''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

LT(s(s(X'')), s(s(Y''))) -> LT(s(X''), s(Y''))

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LT(s(s(X'')), s(s(Y''))) -> LT(s(X''), s(Y''))
one new Dependency Pair is created:

LT(s(s(s(X''''))), s(s(s(Y'''')))) -> LT(s(s(X'''')), s(s(Y'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

LT(s(s(s(X''''))), s(s(s(Y'''')))) -> LT(s(s(X'''')), s(s(Y'''')))

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LT(s(s(s(X''''))), s(s(s(Y'''')))) -> LT(s(s(X'''')), s(s(Y'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(LT(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

APPEND(add(N, X), Y) -> APPEND(X, Y)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

APPEND(add(N, X), Y) -> APPEND(X, Y)
one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(APPEND(x1, x2)) =  1 + x1 POL(add(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

SPLIT(N, add(M, Y)) -> SPLIT(N, Y)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SPLIT(N, add(M, Y)) -> SPLIT(N, Y)
one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(SPLIT(x1, x2)) =  1 + x2 POL(add(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Narrowing Transformation`

Dependency Pairs:

F3(pair(Y, Z), N, X) -> QSORT(Z)
F3(pair(Y, Z), N, X) -> QSORT(Y)
QSORT(add(N, X)) -> F3(split(N, X), N, X)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QSORT(add(N, X)) -> F3(split(N, X), N, X)
two new Dependency Pairs are created:

QSORT(add(N'', nil)) -> F3(pair(nil, nil), N'', nil)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Narrowing Transformation`

Dependency Pairs:

F3(pair(Y, Z), N, X) -> QSORT(Y)
QSORT(add(N'', nil)) -> F3(pair(nil, nil), N'', nil)
F3(pair(Y, Z), N, X) -> QSORT(Z)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F3(pair(Y, Z), N, X) -> QSORT(Z)
QSORT(add(N'', nil)) -> F3(pair(nil, nil), N'', nil)
F3(pair(Y, Z), N, X) -> QSORT(Y)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

one new Dependency Pair is created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Instantiation Transformation`

Dependency Pairs:

F3(pair(Y, Z), N, X) -> QSORT(Z)
QSORT(add(N'', nil)) -> F3(pair(nil, nil), N'', nil)
F3(pair(Y, Z), N, X) -> QSORT(Y)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F3(pair(Y, Z), N, X) -> QSORT(Y)
three new Dependency Pairs are created:

F3(pair(nil, nil), N', nil) -> QSORT(nil)
F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Y')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Instantiation Transformation`

Dependency Pairs:

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Y')
QSORT(add(N'', nil)) -> F3(pair(nil, nil), N'', nil)
F3(pair(Y, Z), N, X) -> QSORT(Z)

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F3(pair(Y, Z), N, X) -> QSORT(Z)
three new Dependency Pairs are created:

F3(pair(nil, nil), N', nil) -> QSORT(nil)
F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')
F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Y')

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')
F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Y')

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Y')
two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F3(pair(Y', Z'), N', add(M''', nil)) -> QSORT(Z')
two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Polynomial Ordering`

Dependency Pairs:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Dependency Graph`

Dependency Pairs:

Rules:

lt(0, s(X)) -> true
lt(s(X), 0) -> false
lt(s(X), s(Y)) -> lt(X, Y)
append(nil, Y) -> Y
split(N, nil) -> pair(nil, nil)
split(N, add(M, Y)) -> f1(split(N, Y), N, M, Y)
f1(pair(X, Z), N, M, Y) -> f2(lt(N, M), N, M, Y, X, Z)
f2(true, N, M, Y, X, Z) -> pair(X, add(M, Z))
f2(false, N, M, Y, X, Z) -> pair(add(M, X), Z)
qsort(nil) -> nil
qsort(add(N, X)) -> f3(split(N, X), N, X)
f3(pair(Y, Z), N, X) -> append(qsort(Y), add(X, qsort(Z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes