YES

The TRS could be proven terminating. The proof took 41693 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (344ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (176ms).
 |    | – Problem 9 was processed with processor PolynomialLinearRange4 (148ms).
 |    |    | – Problem 10 was processed with processor PolynomialLinearRange4 (295ms).
 |    |    |    | – Problem 11 was processed with processor PolynomialLinearRange4 (110ms).
 |    |    |    |    | – Problem 12 was processed with processor PolynomialLinearRange4 (112ms).
 |    |    |    |    |    | – Problem 13 was processed with processor DependencyGraph (1ms).
 |    |    |    |    |    |    | – Problem 14 was processed with processor PolynomialLinearRange4 (31ms).
 |    |    |    |    |    |    |    | – Problem 15 was processed with processor PolynomialLinearRange4 (42ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (34ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 7 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 8 was processed with processor SubtermCriterion (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


The following SCCs where found

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

from#(active(X)) → from#(X)from#(mark(X)) → from#(X)

s#(mark(X)) → s#(X)s#(active(X)) → s#(X)

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

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

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


Polynomial Interpretation

Standard Usable rules

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

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

mark#(s(X)) → active#(s(mark(X)))

Problem 9: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


Polynomial Interpretation

Standard Usable rules

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

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

mark#(from(X)) → mark#(X)active#(from(X)) → mark#(cons(X, from(s(X))))

Problem 10: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


Polynomial Interpretation

Standard Usable rules

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

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

mark#(from(X)) → active#(from(mark(X)))

Problem 11: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


Polynomial Interpretation

Standard Usable rules

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

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

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

Problem 12: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

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

Original Signature

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

Strategy


Polynomial Interpretation

Standard Usable rules

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

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

active#(first(s(X), cons(Y, Z))) → mark#(cons(Y, first(X, Z)))mark#(s(X)) → mark#(X)

Problem 13: DependencyGraph



Dependency Pair Problem

Dependency Pairs

mark#(first(X1, X2)) → mark#(X2)mark#(cons(X1, X2)) → mark#(X1)
mark#(first(X1, X2)) → active#(first(mark(X1), mark(X2)))mark#(first(X1, X2)) → mark#(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


The following SCCs where found

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

Problem 14: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(first(X1, X2)) → mark#(X2)mark#(cons(X1, X2)) → mark#(X1)
mark#(first(X1, X2)) → mark#(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Polynomial Interpretation

There are no usable rules

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

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

Problem 15: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(cons(X1, X2)) → mark#(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Polynomial Interpretation

There are no usable rules

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

mark#(cons(X1, X2)) → mark#(X1)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

from#(active(X)) → from#(X)from#(mark(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

from#(active(X)) → from#(X)from#(mark(X)) → from#(X)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

s#(mark(X)) → s#(X)s#(active(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

s#(mark(X)) → s#(X)s#(active(X)) → s#(X)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(active(X1), X2) → first#(X1, X2)first#(mark(X1), X2) → first#(X1, X2)
first#(X1, mark(X2)) → first#(X1, X2)first#(X1, active(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(X1, active(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons#(X1, active(X2)) → cons#(X1, X2)cons#(mark(X1), X2) → cons#(X1, X2)
cons#(X1, mark(X2)) → cons#(X1, X2)cons#(active(X1), 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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons#(X1, active(X2)) → cons#(X1, X2)cons#(X1, mark(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))))mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)s(mark(X)) → s(X)
s(active(X)) → s(X)cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)from(mark(X)) → from(X)
from(active(X)) → from(X)

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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