Termination Proof using AProVE

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
append(add(N, X), Y) -> add(N, append(X, 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.

     ↳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)
QSORT(add(N, X)) -> SPLIT(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.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

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
append(add(N, X), Y) -> add(N, append(X, 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)))
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
append(add(N, X), Y) -> add(N, append(X, 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)))
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
append(add(N, X), Y) -> add(N, append(X, 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)))
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
append(add(N, X), Y) -> add(N, append(X, 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 could not be shown.
Duration:
0:00 minutes