TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (1228ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 8 was processed with processor ReductionPairSAT (114ms).
 |    |    | – Problem 12 was processed with processor ReductionPairSAT (67ms).
 | – Problem 4 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 9 was processed with processor ReductionPairSAT (47ms).
 |    |    | – Problem 13 was processed with processor ReductionPairSAT (24ms).
 | – Problem 5 was processed with processor ReductionPairSAT (5712ms).
 |    | – Problem 11 was processed with processor ReductionPairSAT (7197ms).
 |    |    | – Problem 15 was processed with processor ReductionPairSAT (5538ms).
 |    |    |    | – Problem 16 remains open; application of the following processors failed [DependencyGraph (103ms), ReductionPairSAT (9532ms), DependencyGraph (100ms), SizeChangePrinciple (timeout)].
 | – Problem 6 was processed with processor SubtermCriterion (3ms).
 | – Problem 7 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 10 was processed with processor ReductionPairSAT (74ms).
 |    |    | – Problem 14 was processed with processor ReductionPairSAT (24ms).

The following open problems remain:

Open Dependency Pair Problem 16

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

The following SCCs where found

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

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

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

Original Signature

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

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: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active = mark < cons# = 0 = s = from = sel = first = cons = nil

Argument Filtering

cons#: collapses to 2
0: all arguments are removed from 0
s: all arguments are removed from s
active: 1
mark: collapses to 1
from: all arguments are removed from from
sel: all arguments are removed from sel
first: 1
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
active: multiset
from: multiset
sel: multiset
first: lexicographic with permutation 1 → 1
cons: lexicographic with permutation 1 → 1 2 → 2
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

cons#(X1, active(X2)) → cons#(X1, X2)

Problem 12: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

cons#(X1, mark(X2)) → cons#(X1, X2)

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

cons# < mark < 0 = s = active = from = sel = first = cons = nil

Argument Filtering

cons#: collapses to 2
0: all arguments are removed from 0
s: all arguments are removed from s
active: all arguments are removed from active
mark: 1
from: 1
sel: 1 2
first: all arguments are removed from first
cons: collapses to 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
active: multiset
mark: multiset
from: lexicographic with permutation 1 → 1
sel: lexicographic with permutation 1 → 1 2 → 2
first: multiset
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

cons#(X1, mark(X2)) → cons#(X1, X2)

Problem 4: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 9: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active = mark < 0 = s = from = sel# = sel = first = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
active: 1
mark: collapses to 1
from: all arguments are removed from from
sel#: 1 2
sel: 1
first: 1
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
active: multiset
from: multiset
sel#: lexicographic with permutation 1 → 1 2 → 2
sel: lexicographic with permutation 1 → 1
first: lexicographic with permutation 1 → 1
cons: multiset
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

sel#(X1, active(X2)) → sel#(X1, X2)

Problem 13: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

sel#(X1, mark(X2)) → sel#(X1, X2)

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

mark < sel# < 0 = s = active = from = sel = first = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
active: all arguments are removed from active
mark: 1
from: all arguments are removed from from
sel#: 2
sel: 1 2
first: 1 2
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
active: multiset
mark: multiset
from: multiset
sel#: multiset
sel: lexicographic with permutation 1 → 1 2 → 2
first: lexicographic with permutation 1 → 1 2 → 2
cons: lexicographic with permutation 1 → 2 2 → 1
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

sel#(X1, mark(X2)) → sel#(X1, X2)

Problem 5: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active < active# < 0 = s = mark = from = sel = mark# = first = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
active: collapses to 1
mark: all arguments are removed from mark
from: all arguments are removed from from
active#: collapses to 1
sel: all arguments are removed from sel
mark#: all arguments are removed from mark#
first: all arguments are removed from first
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
mark: multiset
from: multiset
sel: multiset
mark#: multiset
first: multiset
cons: multiset
nil: multiset

Usable Rules

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

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

mark#(nil) → active#(nil)

Problem 11: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active# < 0 = s = active = mark = from = sel = mark# = first = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
active: collapses to 1
mark: all arguments are removed from mark
from: all arguments are removed from from
active#: collapses to 1
sel: all arguments are removed from sel
mark#: all arguments are removed from mark#
first: all arguments are removed from first
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
mark: multiset
from: multiset
sel: multiset
mark#: multiset
first: multiset
cons: multiset
nil: multiset

Usable Rules

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

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

mark#(0) → active#(0)

Problem 15: ReductionPairSAT

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active < active# < mark = from = sel = mark# = first = nil < s < 0 = cons

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
active: collapses to 1
mark: all arguments are removed from mark
from: all arguments are removed from from
active#: collapses to 1
sel: all arguments are removed from sel
mark#: all arguments are removed from mark#
first: all arguments are removed from first
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
mark: multiset
from: multiset
sel: multiset
mark#: multiset
first: multiset
cons: multiset
nil: multiset

Usable Rules

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

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

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

Problem 6: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

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

Original Signature

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

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 10: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

active = mark < first# = 0 = s = from = sel = first = cons = nil

Argument Filtering

first#: collapses to 2
0: all arguments are removed from 0
s: 1
active: collapses to 1
mark: 1
from: all arguments are removed from from
sel: 1 2
first: all arguments are removed from first
cons: 2
nil: all arguments are removed from nil

Status

0: multiset
s: lexicographic with permutation 1 → 1
mark: multiset
from: multiset
sel: lexicographic with permutation 1 → 1 2 → 2
first: multiset
cons: lexicographic with permutation 2 → 1
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

first#(X1, mark(X2)) → first#(X1, X2)

Problem 14: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy

Function Precedence

first# < active < 0 = s = mark = from = sel = first = cons = nil

Argument Filtering

first#: collapses to 2
0: all arguments are removed from 0
s: all arguments are removed from s
active: 1
mark: all arguments are removed from mark
from: all arguments are removed from from
sel: 1 2
first: all arguments are removed from first
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
active: multiset
mark: multiset
from: multiset
sel: lexicographic with permutation 1 → 1 2 → 2
first: multiset
cons: multiset
nil: multiset

Usable Rules

There are no usable rules.

The dependency pairs and usable rules are stronlgy conservative!

Eliminated dependency pairs

The following dependency pairs (at least) can be eliminated according to the given precedence.

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