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 (1286ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (289ms), PolynomialLinearRange4iUR (2501ms), DependencyGraph (228ms), PolynomialLinearRange8NegiUR (7500ms), DependencyGraph (209ms), ReductionPairSAT (7936ms), DependencyGraph (207ms), ReductionPairSAT (7679ms), DependencyGraph (258ms), ReductionPairSAT (7772ms), DependencyGraph (207ms), SizeChangePrinciple (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 8 was processed with processor ReductionPairSAT (83ms).
 |    |    | – Problem 11 was processed with processor ReductionPairSAT (86ms).
 | – Problem 5 was processed with processor SubtermCriterion (4ms).
 |    | – Problem 9 was processed with processor ReductionPairSAT (86ms).
 |    |    | – Problem 12 was processed with processor ReductionPairSAT (38ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 10 was processed with processor ReductionPairSAT (143ms).
 |    |    | – Problem 13 was processed with processor ReductionPairSAT (29ms).

The following open problems remain:

Open Dependency Pair Problem 2

Dependency Pairs

mark#(0) → active#(0)mark#(cons(X1, X2)) → active#(cons(mark(X1), X2))
mark#(s(X)) → active#(s(mark(X)))mark#(take(X1, X2)) → mark#(X1)
mark#(tt) → active#(tt)mark#(cons(X1, X2)) → mark#(X1)
active#(length(nil)) → mark#(0)mark#(take(X1, X2)) → active#(take(mark(X1), mark(X2)))
active#(and(tt, X)) → mark#(X)active#(take(0, IL)) → mark#(nil)
mark#(and(X1, X2)) → mark#(X1)mark#(nil) → active#(nil)
active#(take(s(M), cons(N, IL))) → mark#(cons(N, take(M, IL)))active#(length(cons(N, L))) → mark#(s(length(L)))
mark#(and(X1, X2)) → active#(and(mark(X1), X2))mark#(take(X1, X2)) → mark#(X2)
mark#(length(X)) → mark#(X)active#(zeros) → mark#(cons(0, zeros))
mark#(s(X)) → mark#(X)mark#(zeros) → active#(zeros)
mark#(length(X)) → active#(length(mark(X)))

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, tt, zeros, take, length, active, mark, nil, and, cons

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

and#(active(X1), X2) → and#(X1, X2)mark#(cons(X1, X2)) → active#(cons(mark(X1), X2))
mark#(take(X1, X2)) → mark#(X1)take#(mark(X1), X2) → take#(X1, X2)
and#(X1, active(X2)) → and#(X1, X2)mark#(tt) → active#(tt)
cons#(mark(X1), X2) → cons#(X1, X2)active#(length(nil)) → mark#(0)
mark#(take(X1, X2)) → active#(take(mark(X1), mark(X2)))active#(take(0, IL)) → mark#(nil)
mark#(s(X)) → s#(mark(X))mark#(nil) → active#(nil)
length#(active(X)) → length#(X)take#(X1, active(X2)) → take#(X1, X2)
mark#(and(X1, X2)) → active#(and(mark(X1), X2))mark#(length(X)) → mark#(X)
active#(take(s(M), cons(N, IL))) → take#(M, IL)active#(take(s(M), cons(N, IL))) → cons#(N, take(M, IL))
cons#(X1, mark(X2)) → cons#(X1, X2)mark#(s(X)) → mark#(X)
mark#(zeros) → active#(zeros)mark#(cons(X1, X2)) → cons#(mark(X1), X2)
and#(X1, mark(X2)) → and#(X1, X2)mark#(length(X)) → length#(mark(X))
length#(mark(X)) → length#(X)mark#(0) → active#(0)
mark#(s(X)) → active#(s(mark(X)))mark#(and(X1, X2)) → and#(mark(X1), X2)
mark#(cons(X1, X2)) → mark#(X1)cons#(active(X1), X2) → cons#(X1, X2)
active#(and(tt, X)) → mark#(X)and#(mark(X1), X2) → and#(X1, X2)
take#(X1, mark(X2)) → take#(X1, X2)active#(length(cons(N, L))) → s#(length(L))
active#(length(cons(N, L))) → length#(L)mark#(take(X1, X2)) → take#(mark(X1), mark(X2))
mark#(and(X1, X2)) → mark#(X1)s#(mark(X)) → s#(X)
cons#(X1, active(X2)) → cons#(X1, X2)active#(take(s(M), cons(N, IL))) → mark#(cons(N, take(M, IL)))
active#(length(cons(N, L))) → mark#(s(length(L)))active#(zeros) → cons#(0, zeros)
mark#(take(X1, X2)) → mark#(X2)active#(zeros) → mark#(cons(0, zeros))
s#(active(X)) → s#(X)mark#(length(X)) → active#(length(mark(X)))
take#(active(X1), X2) → take#(X1, X2)

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

The following SCCs where found

length#(mark(X)) → length#(X)length#(active(X)) → length#(X)
mark#(cons(X1, X2)) → active#(cons(mark(X1), X2))mark#(0) → active#(0)
mark#(take(X1, X2)) → mark#(X1)mark#(s(X)) → active#(s(mark(X)))
mark#(tt) → active#(tt)mark#(cons(X1, X2)) → mark#(X1)
active#(length(nil)) → mark#(0)mark#(take(X1, X2)) → active#(take(mark(X1), mark(X2)))
active#(and(tt, X)) → mark#(X)active#(take(0, IL)) → mark#(nil)
mark#(and(X1, X2)) → mark#(X1)mark#(nil) → active#(nil)
active#(take(s(M), cons(N, IL))) → mark#(cons(N, take(M, IL)))active#(length(cons(N, L))) → mark#(s(length(L)))
mark#(length(X)) → mark#(X)mark#(take(X1, X2)) → mark#(X2)
mark#(and(X1, X2)) → active#(and(mark(X1), X2))active#(zeros) → mark#(cons(0, zeros))
mark#(s(X)) → mark#(X)mark#(zeros) → active#(zeros)
mark#(length(X)) → active#(length(mark(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)
take#(mark(X1), X2) → take#(X1, X2)take#(X1, active(X2)) → take#(X1, X2)
take#(X1, mark(X2)) → take#(X1, X2)take#(active(X1), X2) → take#(X1, X2)
and#(active(X1), X2) → and#(X1, X2)and#(X1, active(X2)) → and#(X1, X2)
and#(mark(X1), X2) → and#(X1, X2)and#(X1, mark(X2)) → and#(X1, X2)

Problem 3: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

length#(mark(X)) → length#(X)length#(active(X)) → length#(X)

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

length#(mark(X)) → length#(X)length#(active(X)) → length#(X)

Problem 4: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 8: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, tt, zeros, take, length, active, mark, nil, and, cons

Strategy

Function Precedence

active < mark < 0 = s = tt = zeros = take = length = and# = and = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: collapses to 1
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: all arguments are removed from take
length: all arguments are removed from length
active: collapses to 1
mark: 1
and#: collapses to 2
and: 1 2
cons: all arguments are removed from cons
nil: all arguments are removed from nil

Status

0: multiset
tt: multiset
zeros: multiset
take: multiset
length: multiset
mark: multiset
and: lexicographic with permutation 1 → 1 2 → 2
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.

and#(X1, mark(X2)) → and#(X1, X2)

Problem 11: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

and#(X1, active(X2)) → and#(X1, X2)

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Function Precedence

0 = s = tt = zeros = take = length = active = mark = and# = and = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: all arguments are removed from take
length: all arguments are removed from length
active: 1
mark: all arguments are removed from mark
and#: collapses to 2
and: 2
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
tt: multiset
zeros: multiset
take: multiset
length: multiset
active: multiset
mark: multiset
and: lexicographic with permutation 2 → 1
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.

and#(X1, active(X2)) → and#(X1, X2)

Problem 5: 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(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, 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 9: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, tt, zeros, take, length, active, mark, nil, and, cons

Strategy

Function Precedence

mark < cons# = 0 = s = tt = zeros = take = length = active = and = cons = nil

Argument Filtering

cons#: 1 2
0: all arguments are removed from 0
s: 1
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: 1 2
length: collapses to 1
active: collapses to 1
mark: 1
and: all arguments are removed from and
cons: 1 2
nil: all arguments are removed from nil

Status

cons#: multiset
0: multiset
s: lexicographic with permutation 1 → 1
tt: multiset
zeros: multiset
take: lexicographic with permutation 1 → 1 2 → 2
mark: lexicographic with permutation 1 → 1
and: multiset
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.

cons#(X1, mark(X2)) → cons#(X1, X2)

Problem 12: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

cons#(X1, active(X2)) → cons#(X1, X2)

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Function Precedence

active < cons# < 0 = s = tt = zeros = take = length = mark = and = cons = nil

Argument Filtering

cons#: collapses to 2
0: all arguments are removed from 0
s: all arguments are removed from s
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: 1
length: 1
active: 1
mark: all arguments are removed from mark
and: all arguments are removed from and
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
tt: multiset
zeros: multiset
take: lexicographic with permutation 1 → 1
length: lexicographic with permutation 1 → 1
active: multiset
mark: multiset
and: multiset
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 6: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, 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

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 10: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, tt, zeros, take, length, active, mark, nil, and, cons

Strategy

Function Precedence

take# < active = mark < 0 = s = tt = zeros = take = length = and = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: all arguments are removed from take
length: all arguments are removed from length
active: collapses to 1
mark: 1
take#: collapses to 2
and: 1 2
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
tt: multiset
zeros: multiset
take: multiset
length: multiset
mark: multiset
and: lexicographic with permutation 1 → 2 2 → 1
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.

take#(X1, mark(X2)) → take#(X1, X2)

Problem 13: ReductionPairSAT

Dependency Pair Problem

Dependency Pairs

take#(X1, active(X2)) → take#(X1, X2)

Rewrite Rules

active(zeros) → mark(cons(0, zeros))active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))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)and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)length(mark(X)) → length(X)
length(active(X)) → length(X)s(mark(X)) → s(X)
s(active(X)) → s(X)take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Original Signature

Termination of terms over the following signature is verified: 0, s, zeros, tt, take, length, active, mark, cons, and, nil

Strategy

Function Precedence

take# < active < 0 = s = tt = zeros = take = length = mark = and = cons = nil

Argument Filtering

0: all arguments are removed from 0
s: all arguments are removed from s
tt: all arguments are removed from tt
zeros: all arguments are removed from zeros
take: all arguments are removed from take
length: all arguments are removed from length
active: 1
mark: 1
take#: collapses to 2
and: all arguments are removed from and
cons: 1 2
nil: all arguments are removed from nil

Status

0: multiset
s: multiset
tt: multiset
zeros: multiset
take: multiset
length: multiset
active: multiset
mark: lexicographic with permutation 1 → 1
and: multiset
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.

take#(X1, active(X2)) → take#(X1, X2)