YES

The TRS could be proven terminating. The proof took 2032 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (400ms).
 | – Problem 2 was processed with processor SubtermCriterion (13ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 9 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (2ms).
 | – Problem 5 was processed with processor PolynomialLinearRange4iUR (1043ms).
 |    | – Problem 10 was processed with processor PolynomialLinearRange4iUR (499ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

proper#(cons(X1, X2)) → proper#(X1)top#(ok(X)) → top#(active(X))
cons#(mark(X1), X2) → cons#(X1, X2)active#(from(X)) → from#(active(X))
active#(first(X1, X2)) → first#(X1, active(X2))cons#(ok(X1), ok(X2)) → cons#(X1, X2)
from#(mark(X)) → from#(X)from#(ok(X)) → from#(X)
top#(ok(X)) → active#(X)active#(first(s(X), cons(Y, Z))) → cons#(Y, first(X, Z))
active#(first(s(X), cons(Y, Z))) → first#(X, Z)active#(cons(X1, X2)) → cons#(active(X1), X2)
active#(from(X)) → cons#(X, from(s(X)))proper#(from(X)) → from#(proper(X))
first#(mark(X1), X2) → first#(X1, X2)top#(mark(X)) → proper#(X)
proper#(from(X)) → proper#(X)first#(X1, mark(X2)) → first#(X1, X2)
active#(first(X1, X2)) → active#(X2)top#(mark(X)) → top#(proper(X))
proper#(cons(X1, X2)) → proper#(X2)active#(from(X)) → active#(X)
proper#(first(X1, X2)) → first#(proper(X1), proper(X2))active#(s(X)) → s#(active(X))
active#(first(X1, X2)) → active#(X1)s#(ok(X)) → s#(X)
proper#(first(X1, X2)) → proper#(X2)s#(mark(X)) → s#(X)
active#(from(X)) → s#(X)proper#(s(X)) → proper#(X)
first#(ok(X1), ok(X2)) → first#(X1, X2)active#(first(X1, X2)) → first#(active(X1), X2)
proper#(cons(X1, X2)) → cons#(proper(X1), proper(X2))active#(s(X)) → active#(X)
proper#(s(X)) → s#(proper(X))proper#(first(X1, X2)) → proper#(X1)
active#(from(X)) → from#(s(X))active#(cons(X1, X2)) → active#(X1)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


The following SCCs where found

cons#(mark(X1), X2) → cons#(X1, X2)cons#(ok(X1), ok(X2)) → cons#(X1, X2)

proper#(first(X1, X2)) → proper#(X2)proper#(s(X)) → proper#(X)
proper#(cons(X1, X2)) → proper#(X1)proper#(cons(X1, X2)) → proper#(X2)
proper#(first(X1, X2)) → proper#(X1)proper#(from(X)) → proper#(X)

from#(mark(X)) → from#(X)from#(ok(X)) → from#(X)

top#(mark(X)) → top#(proper(X))top#(ok(X)) → top#(active(X))

s#(mark(X)) → s#(X)s#(ok(X)) → s#(X)

active#(first(X1, X2)) → active#(X2)active#(from(X)) → active#(X)
active#(s(X)) → active#(X)active#(first(X1, X2)) → active#(X1)
active#(cons(X1, X2)) → active#(X1)

first#(ok(X1), ok(X2)) → first#(X1, X2)first#(mark(X1), X2) → first#(X1, X2)
first#(X1, mark(X2)) → first#(X1, X2)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

active#(first(X1, X2)) → active#(X2)active#(from(X)) → active#(X)
active#(s(X)) → active#(X)active#(first(X1, X2)) → active#(X1)
active#(cons(X1, X2)) → active#(X1)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

active#(first(X1, X2)) → active#(X2)active#(from(X)) → active#(X)
active#(s(X)) → active#(X)active#(first(X1, X2)) → active#(X1)
active#(cons(X1, X2)) → active#(X1)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(ok(X1), ok(X2)) → first#(X1, X2)first#(mark(X1), X2) → first#(X1, X2)
first#(X1, mark(X2)) → first#(X1, X2)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(ok(X1), ok(X2)) → first#(X1, X2)first#(mark(X1), X2) → first#(X1, X2)

Problem 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(X1, mark(X2)) → first#(X1, X2)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, ok, mark, proper, from, first, top, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(X1, mark(X2)) → first#(X1, X2)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

proper#(first(X1, X2)) → proper#(X2)proper#(s(X)) → proper#(X)
proper#(cons(X1, X2)) → proper#(X1)proper#(cons(X1, X2)) → proper#(X2)
proper#(first(X1, X2)) → proper#(X1)proper#(from(X)) → proper#(X)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

proper#(first(X1, X2)) → proper#(X2)proper#(s(X)) → proper#(X)
proper#(cons(X1, X2)) → proper#(X1)proper#(cons(X1, X2)) → proper#(X2)
proper#(first(X1, X2)) → proper#(X1)proper#(from(X)) → proper#(X)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(mark(X)) → top#(proper(X))top#(ok(X)) → top#(active(X))

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Polynomial Interpretation

Improved Usable rules

active(s(X)) → s(active(X))s(ok(X)) → ok(s(X))
first(mark(X1), X2) → mark(first(X1, X2))cons(mark(X1), X2) → mark(cons(X1, X2))
active(first(0, X)) → mark(nil)from(ok(X)) → ok(from(X))
first(X1, mark(X2)) → mark(first(X1, X2))active(from(X)) → from(active(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))from(mark(X)) → mark(from(X))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))active(cons(X1, X2)) → cons(active(X1), X2)
s(mark(X)) → mark(s(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(nil) → ok(nil)active(first(X1, X2)) → first(active(X1), X2)
proper(s(X)) → s(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
proper(0) → ok(0)active(from(X)) → mark(cons(X, from(s(X))))
proper(first(X1, X2)) → first(proper(X1), proper(X2))proper(from(X)) → from(proper(X))
active(first(X1, X2)) → first(X1, active(X2))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

top#(mark(X)) → top#(proper(X))

Problem 10: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(ok(X)) → top#(active(X))

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, ok, mark, proper, from, first, top, cons, nil

Strategy


Polynomial Interpretation

Improved Usable rules

active(s(X)) → s(active(X))s(ok(X)) → ok(s(X))
first(mark(X1), X2) → mark(first(X1, X2))cons(mark(X1), X2) → mark(cons(X1, X2))
from(ok(X)) → ok(from(X))active(first(0, X)) → mark(nil)
first(X1, mark(X2)) → mark(first(X1, X2))active(from(X)) → from(active(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))from(mark(X)) → mark(from(X))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))s(mark(X)) → mark(s(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(first(X1, X2)) → first(active(X1), X2)
first(ok(X1), ok(X2)) → ok(first(X1, X2))active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(X1, active(X2))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

top#(ok(X)) → top#(active(X))

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons#(mark(X1), X2) → cons#(X1, X2)cons#(ok(X1), ok(X2)) → cons#(X1, X2)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

cons#(mark(X1), X2) → cons#(X1, X2)cons#(ok(X1), ok(X2)) → cons#(X1, X2)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

s#(mark(X)) → s#(X)s#(ok(X)) → s#(X)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

s#(mark(X)) → s#(X)s#(ok(X)) → s#(X)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

from#(mark(X)) → from#(X)from#(ok(X)) → from#(X)

Rewrite Rules

active(first(0, X)) → mark(nil)active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Original Signature

Termination of terms over the following signature is verified: 0, s, active, mark, ok, from, proper, first, nil, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

from#(mark(X)) → from#(X)from#(ok(X)) → from#(X)