TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60234 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (346ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (121ms).
 |    | – Problem 5 was processed with processor DependencyGraph (3ms).
 |    |    | – Problem 7 was processed with processor PolynomialLinearRange4 (18ms).
 |    |    |    | – Problem 10 was processed with processor DependencyGraph (1ms).
 |    |    |    |    | – Problem 11 was processed with processor PolynomialLinearRange4 (13ms).
 |    |    |    |    | – Problem 12 was processed with processor PolynomialLinearRange4 (8ms).
 |    |    | – Problem 8 was processed with processor PolynomialLinearRange4 (8ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4 (505ms).
 |    | – Problem 6 was processed with processor DependencyGraph (1ms).
 |    |    | – Problem 9 was processed with processor PolynomialLinearRange4 (12ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (0ms), DependencyGraph (0ms), PolynomialOrderingProcessor (0ms), DependencyGraph (1ms), PolynomialLinearRange4 (129ms), DependencyGraph (1ms), PolynomialLinearRange4 (119ms), DependencyGraph (1ms), PolynomialLinearRange4 (124ms), DependencyGraph (1ms), PolynomialLinearRange4 (129ms), DependencyGraph (0ms), ReductionPairSAT (756ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].

The following open problems remain:

Open Dependency Pair Problem 4

Dependency Pairs

uLength#(tt, L) → length#(L)length#(cons(N, L)) → uLength#(and(isNat(N), isNatList(L)), L)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

isNat#(s(N)) → isNat#(N)isNat#(length(L)) → isNatList#(L)
isNatIList#(cons(N, IL)) → and#(isNat(N), isNatIList(IL))take#(s(M), cons(N, IL)) → uTake2#(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2#(tt, M, N, IL) → T(N)take#(s(M), cons(N, IL)) → isNatIList#(IL)
isNatList#(cons(N, L)) → isNat#(N)uLength#(tt, L) → length#(L)
T(take(x_1, x_2)) → T(x_2)length#(cons(N, L)) → and#(isNat(N), isNatList(L))
isNatIList#(cons(N, IL)) → isNat#(N)T(zeros) → zeros#
take#(s(M), cons(N, IL)) → and#(isNat(N), isNatIList(IL))take#(s(M), cons(N, IL)) → isNat#(N)
uLength#(tt, L) → T(L)isNatList#(take(N, IL)) → isNatIList#(IL)
take#(s(M), cons(N, IL)) → isNat#(M)isNatIList#(cons(N, IL)) → isNatIList#(IL)
isNatIList#(IL) → isNatList#(IL)T(take(x_1, x_2)) → T(x_1)
length#(cons(N, L)) → uLength#(and(isNat(N), isNatList(L)), L)take#(s(M), cons(N, IL)) → and#(isNat(M), and(isNat(N), isNatIList(IL)))
isNatList#(cons(N, L)) → isNatList#(L)isNatList#(take(N, IL)) → and#(isNat(N), isNatIList(IL))
take#(0, IL) → isNatIList#(IL)isNatList#(cons(N, L)) → and#(isNat(N), isNatList(L))
length#(cons(N, L)) → isNat#(N)isNatList#(take(N, IL)) → isNat#(N)
take#(0, IL) → uTake1#(isNatIList(IL))T(take(M, IL)) → take#(M, IL)
length#(cons(N, L)) → isNatList#(L)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

The following SCCs where found

isNatList#(cons(N, L)) → isNat#(N)isNat#(s(N)) → isNat#(N)
isNat#(length(L)) → isNatList#(L)isNatIList#(cons(N, IL)) → isNat#(N)
isNatIList#(IL) → isNatList#(IL)isNatList#(take(N, IL)) → isNat#(N)
isNatList#(cons(N, L)) → isNatList#(L)isNatList#(take(N, IL)) → isNatIList#(IL)
isNatIList#(cons(N, IL)) → isNatIList#(IL)
uLength#(tt, L) → length#(L)length#(cons(N, L)) → uLength#(and(isNat(N), isNatList(L)), L)
T(take(x_1, x_2)) → T(x_2)take#(s(M), cons(N, IL)) → uTake2#(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
T(take(x_1, x_2)) → T(x_1)T(take(M, IL)) → take#(M, IL)
uTake2#(tt, M, N, IL) → T(N)

Problem 2: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

isNatList#(cons(N, L)) → isNat#(N)isNat#(s(N)) → isNat#(N)
isNat#(length(L)) → isNatList#(L)isNatIList#(cons(N, IL)) → isNat#(N)
isNatIList#(IL) → isNatList#(IL)isNatList#(take(N, IL)) → isNat#(N)
isNatList#(cons(N, L)) → isNatList#(L)isNatList#(take(N, IL)) → isNatIList#(IL)
isNatIList#(cons(N, IL)) → isNatIList#(IL)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

isNatList#(cons(N, L)) → isNat#(N)isNatIList#(cons(N, IL)) → isNat#(N)
isNatList#(take(N, IL)) → isNat#(N)

Problem 5: DependencyGraph

Dependency Pair Problem

Dependency Pairs

isNat#(s(N)) → isNat#(N)isNat#(length(L)) → isNatList#(L)
isNatIList#(IL) → isNatList#(IL)isNatList#(cons(N, L)) → isNatList#(L)
isNatList#(take(N, IL)) → isNatIList#(IL)isNatIList#(cons(N, IL)) → isNatIList#(IL)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

The following SCCs where found

isNat#(s(N)) → isNat#(N)
isNatIList#(IL) → isNatList#(IL)isNatList#(cons(N, L)) → isNatList#(L)
isNatList#(take(N, IL)) → isNatIList#(IL)isNatIList#(cons(N, IL)) → isNatIList#(IL)

Problem 7: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

isNatIList#(IL) → isNatList#(IL)isNatList#(cons(N, L)) → isNatList#(L)
isNatList#(take(N, IL)) → isNatIList#(IL)isNatIList#(cons(N, IL)) → isNatIList#(IL)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

isNatIList#(IL) → isNatList#(IL)

Problem 10: DependencyGraph

Dependency Pair Problem

Dependency Pairs

isNatList#(cons(N, L)) → isNatList#(L)isNatList#(take(N, IL)) → isNatIList#(IL)
isNatIList#(cons(N, IL)) → isNatIList#(IL)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

The following SCCs where found

isNatList#(cons(N, L)) → isNatList#(L)
isNatIList#(cons(N, IL)) → isNatIList#(IL)

Problem 11: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

isNatList#(cons(N, L)) → isNatList#(L)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

isNatList#(cons(N, L)) → isNatList#(L)

Problem 12: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

isNatIList#(cons(N, IL)) → isNatIList#(IL)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

isNatIList#(cons(N, IL)) → isNatIList#(IL)

Problem 8: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

isNat#(s(N)) → isNat#(N)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

isNat#(s(N)) → isNat#(N)

Problem 3: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

T(take(x_1, x_2)) → T(x_2)take#(s(M), cons(N, IL)) → uTake2#(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
T(take(x_1, x_2)) → T(x_1)T(take(M, IL)) → take#(M, IL)
uTake2#(tt, M, N, IL) → T(N)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, nil, cons

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

Polynomial Interpretation

Standard Usable rules

isNat(length(L)) → isNatList(L)take(0, IL) → uTake1(isNatIList(IL))
uLength(tt, L) → s(length(L))isNat(0) → tt
isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
zeros → cons(0, zeros)isNatIList(IL) → isNatList(IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))and(tt, T) → T
length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)isNat(s(N)) → isNat(N)
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))

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

take#(s(M), cons(N, IL)) → uTake2#(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)

Problem 6: DependencyGraph

Dependency Pair Problem

Dependency Pairs

T(take(x_1, x_2)) → T(x_2)T(take(x_1, x_2)) → T(x_1)
T(take(M, IL)) → take#(M, IL)uTake2#(tt, M, N, IL) → T(N)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

The following SCCs where found

T(take(x_1, x_2)) → T(x_2)T(take(x_1, x_2)) → T(x_1)

Problem 9: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

T(take(x_1, x_2)) → T(x_2)T(take(x_1, x_2)) → T(x_1)

Rewrite Rules

and(tt, T) → TisNatIList(IL) → isNatList(IL)
isNat(0) → ttisNat(s(N)) → isNat(N)
isNat(length(L)) → isNatList(L)isNatIList(zeros) → tt
isNatIList(cons(N, IL)) → and(isNat(N), isNatIList(IL))isNatList(nil) → tt
isNatList(cons(N, L)) → and(isNat(N), isNatList(L))isNatList(take(N, IL)) → and(isNat(N), isNatIList(IL))
zeros → cons(0, zeros)take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → niltake(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(IL))), M, N, IL)
uTake2(tt, M, N, IL) → cons(N, take(M, IL))length(cons(N, L)) → uLength(and(isNat(N), isNatList(L)), L)
uLength(tt, L) → s(length(L))

Original Signature

Termination of terms over the following signature is verified: isNatIList, uLength, and, isNat, uTake1, uTake2, 0, isNatList, s, tt, zeros, take, length, cons, nil

Strategy

Context-sensitive strategy:
μ(isNatList#) = μ(zeros#) = μ(isNat) = μ(T) = μ(isNatIList) = μ(isNatIList#) = μ(0) = μ(isNatList) = μ(tt) = μ(zeros) = μ(isNat#) = μ(nil) = ∅
μ(uLength#) = μ(length#) = μ(uTake2#) = μ(length) = μ(cons) = μ(uLength) = μ(uTake1#) = μ(uTake1) = μ(uTake2) = μ(s) = {1}
μ(and#) = μ(take#) = μ(and) = μ(take) = {1, 2}

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:

T(take(x_1, x_2)) → T(x_2)T(take(x_1, x_2)) → T(x_1)