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)))

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`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(f_1(x1, x2, x3, x4)) =  0 POL(false) =  0 POL(lt(x1, x2)) =  0 POL(true) =  0 POL(append(x1, x2)) =  x2 POL(add(x1, x2)) =  0 POL(f_3(x1, x2, x3)) =  0 POL(qsort(x1)) =  0 POL(0) =  0 POL(pair(x1, x2)) =  0 POL(LT(x1, x2)) =  x1 POL(split(x1, x2)) =  0 POL(f_2(x1, x2, x3, x4, x5, x6)) =  0 POL(nil) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`

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)))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(f_1(x1, x2, x3, x4)) =  1 + x1 POL(false) =  0 POL(lt(x1, x2)) =  0 POL(true) =  0 POL(append(x1, x2)) =  x1 + x2 POL(APPEND(x1, x2)) =  1 + x1 POL(add(x1, x2)) =  1 + x2 POL(f_3(x1, x2, x3)) =  1 + x1 POL(qsort(x1)) =  x1 POL(0) =  0 POL(pair(x1, x2)) =  x1 + x2 POL(split(x1, x2)) =  x2 POL(f_2(x1, x2, x3, x4, x5, x6)) =  1 + x5 + x6 POL(nil) =  0 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`

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)))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(f_1(x1, x2, x3, x4)) =  1 + x1 POL(false) =  0 POL(SPLIT(x1, x2)) =  1 + x2 POL(lt(x1, x2)) =  0 POL(true) =  0 POL(append(x1, x2)) =  x1 + x2 POL(add(x1, x2)) =  1 + x2 POL(f_3(x1, x2, x3)) =  1 + x1 POL(qsort(x1)) =  x1 POL(0) =  0 POL(pair(x1, x2)) =  x1 + x2 POL(split(x1, x2)) =  x2 POL(f_2(x1, x2, x3, x4, x5, x6)) =  1 + x5 + x6 POL(nil) =  0 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`

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)))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polynomial Ordering`

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)))

The following dependency pairs can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

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)) =  1 + x1 POL(lt(x1, x2)) =  0 POL(true) =  0 POL(append(x1, x2)) =  x1 + x2 POL(add(x1, x2)) =  1 + x2 POL(f_3(x1, x2, x3)) =  1 + x1 POL(qsort(x1)) =  x1 POL(0) =  0 POL(pair(x1, x2)) =  x1 + x2 POL(split(x1, x2)) =  x2 POL(f_2(x1, x2, x3, x4, x5, x6)) =  1 + x5 + x6 POL(nil) =  0 POL(s(x1)) =  0 POL(QSORT(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Dependency Graph`

Dependency Pair:

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)))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes