YES

The TRS could be proven terminating. The proof took 42420 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (673ms).
 | – Problem 2 was processed with processor SubtermCriterion (8ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 9 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor PolynomialLinearRange4 (236ms).
 |    | – Problem 12 was processed with processor PolynomialLinearRange4 (362ms).
 |    |    | – Problem 13 was processed with processor PolynomialLinearRange4 (199ms).
 |    |    |    | – Problem 14 was processed with processor PolynomialLinearRange4 (161ms).
 |    |    |    |    | – Problem 15 was processed with processor PolynomialLinearRange4 (161ms).
 |    |    |    |    |    | – Problem 16 was processed with processor PolynomialLinearRange4 (126ms).
 |    |    |    |    |    |    | – Problem 17 was processed with processor PolynomialLinearRange4 (52ms).
 |    |    |    |    |    |    |    | – Problem 18 was processed with processor DependencyGraph (1ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 10 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 11 was processed with processor SubtermCriterion (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)mark#(take(X1, X2))mark#(X1)
mark#(eq(X1, X2))active#(eq(X1, X2))active#(length(nil))mark#(0)
mark#(take(X1, X2))active#(take(mark(X1), mark(X2)))active#(eq(s(X), s(Y)))eq#(X, Y)
active#(take(s(X), cons(Y, L)))mark#(cons(Y, take(X, L)))length#(active(X))length#(X)
mark#(eq(X1, X2))eq#(X1, X2)active#(take(s(X), cons(Y, L)))take#(X, L)
inf#(active(X))inf#(X)length#(mark(X))length#(X)
mark#(true)active#(true)active#(length(cons(X, L)))length#(L)
take#(X1, mark(X2))take#(X1, X2)mark#(inf(X))inf#(mark(X))
mark#(s(X))active#(s(X))active#(take(s(X), cons(Y, L)))cons#(Y, take(X, L))
cons#(X1, active(X2))cons#(X1, X2)eq#(mark(X1), X2)eq#(X1, X2)
active#(inf(X))cons#(X, inf(s(X)))mark#(false)active#(false)
take#(mark(X1), X2)take#(X1, X2)active#(length(cons(X, L)))mark#(s(length(L)))
cons#(mark(X1), X2)cons#(X1, X2)mark#(cons(X1, X2))cons#(X1, X2)
eq#(X1, mark(X2))eq#(X1, X2)active#(take(0, X))mark#(nil)
eq#(active(X1), X2)eq#(X1, X2)eq#(X1, active(X2))eq#(X1, X2)
active#(eq(s(X), s(Y)))mark#(eq(X, Y))active#(eq(X, Y))mark#(false)
active#(inf(X))s#(X)mark#(nil)active#(nil)
take#(X1, active(X2))take#(X1, X2)mark#(length(X))mark#(X)
mark#(inf(X))active#(inf(mark(X)))cons#(X1, mark(X2))cons#(X1, X2)
mark#(0)active#(0)active#(inf(X))mark#(cons(X, inf(s(X))))
mark#(length(X))length#(mark(X))active#(inf(X))inf#(s(X))
active#(length(cons(X, L)))s#(length(L))cons#(active(X1), X2)cons#(X1, X2)
mark#(cons(X1, X2))active#(cons(X1, X2))mark#(take(X1, X2))take#(mark(X1), mark(X2))
mark#(s(X))s#(X)s#(mark(X))s#(X)
mark#(take(X1, X2))mark#(X2)s#(active(X))s#(X)
inf#(mark(X))inf#(X)mark#(length(X))active#(length(mark(X)))
active#(eq(0, 0))mark#(true)take#(active(X1), X2)take#(X1, X2)

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


The following SCCs where found

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

inf#(mark(X)) → inf#(X)inf#(active(X)) → inf#(X)

mark#(inf(X)) → mark#(X)active#(inf(X)) → mark#(cons(X, inf(s(X))))
mark#(take(X1, X2)) → mark#(X1)active#(length(cons(X, L))) → mark#(s(length(L)))
mark#(eq(X1, X2)) → active#(eq(X1, X2))mark#(take(X1, X2)) → active#(take(mark(X1), mark(X2)))
mark#(s(X)) → active#(s(X))mark#(cons(X1, X2)) → active#(cons(X1, X2))
active#(take(s(X), cons(Y, L))) → mark#(cons(Y, take(X, L)))active#(eq(s(X), s(Y))) → mark#(eq(X, Y))
mark#(length(X)) → mark#(X)mark#(take(X1, X2)) → mark#(X2)
mark#(inf(X)) → active#(inf(mark(X)))mark#(length(X)) → active#(length(mark(X)))

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

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

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, 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 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

inf#(mark(X))inf#(X)inf#(active(X))inf#(X)

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

inf#(mark(X))inf#(X)inf#(active(X))inf#(X)

Problem 4: 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(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, 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 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, 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 6: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)active#(inf(X))mark#(cons(X, inf(s(X))))
active#(length(cons(X, L)))mark#(s(length(L)))mark#(take(X1, X2))mark#(X1)
mark#(eq(X1, X2))active#(eq(X1, X2))mark#(take(X1, X2))active#(take(mark(X1), mark(X2)))
mark#(s(X))active#(s(X))mark#(cons(X1, X2))active#(cons(X1, X2))
active#(take(s(X), cons(Y, L)))mark#(cons(Y, take(X, L)))active#(eq(s(X), s(Y)))mark#(eq(X, Y))
mark#(take(X1, X2))mark#(X2)mark#(length(X))mark#(X)
mark#(inf(X))active#(inf(mark(X)))mark#(length(X))active#(length(mark(X)))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))length(mark(X))length(X)
mark(s(X))active(s(X))inf(mark(X))inf(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)length(active(X))length(X)
s(mark(X))s(X)eq(mark(X1), X2)eq(X1, X2)
take(active(X1), X2)take(X1, X2)active(eq(0, 0))mark(true)

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

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

Problem 12: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)active#(inf(X))mark#(cons(X, inf(s(X))))
active#(length(cons(X, L)))mark#(s(length(L)))mark#(take(X1, X2))mark#(X1)
mark#(eq(X1, X2))active#(eq(X1, X2))mark#(take(X1, X2))active#(take(mark(X1), mark(X2)))
mark#(cons(X1, X2))active#(cons(X1, X2))active#(take(s(X), cons(Y, L)))mark#(cons(Y, take(X, L)))
active#(eq(s(X), s(Y)))mark#(eq(X, Y))mark#(take(X1, X2))mark#(X2)
mark#(length(X))mark#(X)mark#(inf(X))active#(inf(mark(X)))
mark#(length(X))active#(length(mark(X)))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))length(mark(X))length(X)
mark(s(X))active(s(X))inf(mark(X))inf(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)length(active(X))length(X)
s(mark(X))s(X)eq(mark(X1), X2)eq(X1, X2)
take(active(X1), X2)take(X1, X2)active(eq(0, 0))mark(true)

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

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

Problem 13: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)active#(take(s(X), cons(Y, L)))mark#(cons(Y, take(X, L)))
active#(inf(X))mark#(cons(X, inf(s(X))))active#(eq(s(X), s(Y)))mark#(eq(X, Y))
active#(length(cons(X, L)))mark#(s(length(L)))mark#(length(X))mark#(X)
mark#(eq(X1, X2))active#(eq(X1, X2))mark#(inf(X))active#(inf(mark(X)))
mark#(length(X))active#(length(mark(X)))mark#(cons(X1, X2))active#(cons(X1, X2))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))length(mark(X))length(X)
mark(s(X))active(s(X))inf(mark(X))inf(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)length(active(X))length(X)
s(mark(X))s(X)eq(mark(X1), X2)eq(X1, X2)
take(active(X1), X2)take(X1, X2)active(eq(0, 0))mark(true)

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

active#(take(s(X), cons(Y, L)))mark#(cons(Y, take(X, L)))

Problem 14: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)active#(inf(X))mark#(cons(X, inf(s(X))))
active#(length(cons(X, L)))mark#(s(length(L)))active#(eq(s(X), s(Y)))mark#(eq(X, Y))
mark#(length(X))mark#(X)mark#(eq(X1, X2))active#(eq(X1, X2))
mark#(inf(X))active#(inf(mark(X)))mark#(length(X))active#(length(mark(X)))
mark#(cons(X1, X2))active#(cons(X1, X2))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))mark(s(X))active(s(X))
inf(mark(X))inf(X)length(mark(X))length(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)s(mark(X))s(X)
length(active(X))length(X)take(active(X1), X2)take(X1, X2)
eq(mark(X1), X2)eq(X1, X2)active(eq(0, 0))mark(true)

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(X1, X2))

Problem 15: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

mark#(inf(X))mark#(X)active#(inf(X))mark#(cons(X, inf(s(X))))
active#(eq(s(X), s(Y)))mark#(eq(X, Y))active#(length(cons(X, L)))mark#(s(length(L)))
mark#(length(X))mark#(X)mark#(eq(X1, X2))active#(eq(X1, X2))
mark#(inf(X))active#(inf(mark(X)))mark#(length(X))active#(length(mark(X)))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))mark(s(X))active(s(X))
inf(mark(X))inf(X)length(mark(X))length(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)s(mark(X))s(X)
length(active(X))length(X)take(active(X1), X2)take(X1, X2)
eq(mark(X1), X2)eq(X1, X2)active(eq(0, 0))mark(true)

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

mark#(inf(X))mark#(X)active#(length(cons(X, L)))mark#(s(length(L)))
mark#(length(X))mark#(X)

Problem 16: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

active#(inf(X))mark#(cons(X, inf(s(X))))active#(eq(s(X), s(Y)))mark#(eq(X, Y))
mark#(eq(X1, X2))active#(eq(X1, X2))mark#(inf(X))active#(inf(mark(X)))
mark#(length(X))active#(length(mark(X)))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


Polynomial Interpretation

Standard Usable rules

active(take(0, X))mark(nil)cons(active(X1), X2)cons(X1, X2)
active(length(cons(X, L)))mark(s(length(L)))inf(active(X))inf(X)
mark(cons(X1, X2))active(cons(X1, X2))mark(s(X))active(s(X))
inf(mark(X))inf(X)length(mark(X))length(X)
active(length(nil))mark(0)cons(X1, mark(X2))cons(X1, X2)
mark(true)active(true)eq(X1, mark(X2))eq(X1, X2)
mark(nil)active(nil)mark(0)active(0)
s(active(X))s(X)take(X1, active(X2))take(X1, X2)
cons(X1, active(X2))cons(X1, X2)active(eq(X, Y))mark(false)
active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))mark(eq(X1, X2))active(eq(X1, X2))
mark(length(X))active(length(mark(X)))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(inf(X))active(inf(mark(X)))active(inf(X))mark(cons(X, inf(s(X))))
take(mark(X1), X2)take(X1, X2)eq(X1, active(X2))eq(X1, X2)
take(X1, mark(X2))take(X1, X2)active(eq(s(X), s(Y)))mark(eq(X, Y))
mark(false)active(false)cons(mark(X1), X2)cons(X1, X2)
eq(active(X1), X2)eq(X1, X2)s(mark(X))s(X)
length(active(X))length(X)take(active(X1), X2)take(X1, X2)
eq(mark(X1), X2)eq(X1, X2)active(eq(0, 0))mark(true)

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

active#(inf(X))mark#(cons(X, inf(s(X))))mark#(inf(X))active#(inf(mark(X)))
mark#(length(X))active#(length(mark(X)))

Problem 17: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

active#(eq(s(X), s(Y)))mark#(eq(X, Y))mark#(eq(X1, X2))active#(eq(X1, X2))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Polynomial Interpretation

Standard Usable rules

eq(X1, active(X2))eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(mark(X1), X2)eq(X1, X2)

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

mark#(eq(X1, X2))active#(eq(X1, X2))

Problem 18: DependencyGraph



Dependency Pair Problem

Dependency Pairs

active#(eq(s(X), s(Y)))mark#(eq(X, Y))

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


There are no SCCs!

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, 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 10: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

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)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(X1, active(X2))eq#(X1, X2)eq#(active(X1), X2)eq#(X1, X2)
eq#(mark(X1), X2)eq#(X1, X2)eq#(X1, mark(X2))eq#(X1, X2)

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, inf, length, true, active, false, mark, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(active(X1), X2)eq#(X1, X2)eq#(mark(X1), X2)eq#(X1, X2)

Problem 11: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(X1, active(X2))eq#(X1, X2)eq#(X1, mark(X2))eq#(X1, X2)

Rewrite Rules

active(eq(0, 0))mark(true)active(eq(s(X), s(Y)))mark(eq(X, Y))
active(eq(X, Y))mark(false)active(inf(X))mark(cons(X, inf(s(X))))
active(take(0, X))mark(nil)active(take(s(X), cons(Y, L)))mark(cons(Y, take(X, L)))
active(length(nil))mark(0)active(length(cons(X, L)))mark(s(length(L)))
mark(eq(X1, X2))active(eq(X1, X2))mark(0)active(0)
mark(true)active(true)mark(s(X))active(s(X))
mark(false)active(false)mark(inf(X))active(inf(mark(X)))
mark(cons(X1, X2))active(cons(X1, X2))mark(take(X1, X2))active(take(mark(X1), mark(X2)))
mark(nil)active(nil)mark(length(X))active(length(mark(X)))
eq(mark(X1), X2)eq(X1, X2)eq(X1, mark(X2))eq(X1, X2)
eq(active(X1), X2)eq(X1, X2)eq(X1, active(X2))eq(X1, X2)
s(mark(X))s(X)s(active(X))s(X)
inf(mark(X))inf(X)inf(active(X))inf(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)
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)
length(mark(X))length(X)length(active(X))length(X)

Original Signature

Termination of terms over the following signature is verified: 0, s, take, length, inf, false, active, true, mark, nil, cons, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(X1, active(X2))eq#(X1, X2)eq#(X1, mark(X2))eq#(X1, X2)