YES

The TRS could be proven terminating. The proof took 2272 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (525ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (1068ms).
 |    | – Problem 11 was processed with processor PolynomialLinearRange4iUR (573ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor SubtermCriterion (0ms).
 | – 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 9 was processed with processor SubtermCriterion (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

proper#(cons(X1, X2))proper#(X1)top#(ok(X))top#(active(X))
cons#(mark(X1), X2)cons#(X1, X2)active#(2nd(X))2nd#(active(X))
2nd#(mark(X))2nd#(X)active#(from(X))from#(active(X))
cons#(ok(X1), ok(X2))cons#(X1, X2)active#(2nd(cons(X, X1)))2nd#(cons1(X, X1))
from#(mark(X))from#(X)from#(ok(X))from#(X)
top#(ok(X))active#(X)active#(cons1(X1, X2))cons1#(active(X1), X2)
active#(cons(X1, X2))cons#(active(X1), X2)2nd#(ok(X))2nd#(X)
proper#(2nd(X))proper#(X)active#(2nd(cons(X, X1)))cons1#(X, X1)
proper#(cons1(X1, X2))proper#(X1)active#(from(X))cons#(X, from(s(X)))
proper#(from(X))from#(proper(X))cons1#(X1, mark(X2))cons1#(X1, X2)
top#(mark(X))proper#(X)proper#(from(X))proper#(X)
active#(2nd(X))active#(X)top#(mark(X))top#(proper(X))
proper#(cons(X1, X2))proper#(X2)active#(from(X))active#(X)
cons1#(ok(X1), ok(X2))cons1#(X1, X2)active#(cons1(X1, X2))cons1#(X1, active(X2))
proper#(2nd(X))2nd#(proper(X))active#(cons1(X1, X2))active#(X2)
active#(s(X))s#(active(X))s#(ok(X))s#(X)
s#(mark(X))s#(X)active#(from(X))s#(X)
proper#(s(X))proper#(X)proper#(cons1(X1, X2))proper#(X2)
cons1#(mark(X1), X2)cons1#(X1, X2)proper#(cons(X1, X2))cons#(proper(X1), proper(X2))
proper#(cons1(X1, X2))cons1#(proper(X1), proper(X2))active#(s(X))active#(X)
proper#(s(X))s#(proper(X))active#(cons1(X1, X2))active#(X1)
active#(from(X))from#(s(X))active#(cons(X1, X2))active#(X1)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


The following SCCs where found

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

proper#(s(X)) → proper#(X)proper#(cons1(X1, X2)) → proper#(X2)
proper#(2nd(X)) → proper#(X)proper#(cons(X1, X2)) → proper#(X1)
proper#(cons(X1, X2)) → proper#(X2)proper#(cons1(X1, X2)) → proper#(X1)
proper#(from(X)) → proper#(X)

active#(from(X)) → active#(X)active#(s(X)) → active#(X)
active#(cons1(X1, X2)) → active#(X1)active#(cons1(X1, X2)) → active#(X2)
active#(2nd(X)) → active#(X)active#(cons(X1, X2)) → active#(X1)

from#(mark(X)) → from#(X)from#(ok(X)) → from#(X)

top#(mark(X)) → top#(proper(X))top#(ok(X)) → top#(active(X))

cons1#(mark(X1), X2) → cons1#(X1, X2)cons1#(ok(X1), ok(X2)) → cons1#(X1, X2)
cons1#(X1, mark(X2)) → cons1#(X1, X2)

s#(mark(X)) → s#(X)s#(ok(X)) → s#(X)

2nd#(ok(X)) → 2nd#(X)2nd#(mark(X)) → 2nd#(X)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(mark(X))top#(proper(X))top#(ok(X))top#(active(X))

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Polynomial Interpretation

Improved Usable rules

active(s(X))s(active(X))cons(mark(X1), X2)mark(cons(X1, X2))
active(from(X))from(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))active(cons1(X1, X2))cons1(X1, active(X2))
s(mark(X))mark(s(X))2nd(ok(X))ok(2nd(X))
proper(s(X))s(proper(X))active(from(X))mark(cons(X, from(s(X))))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))2nd(mark(X))mark(2nd(X))
s(ok(X))ok(s(X))cons1(X1, mark(X2))mark(cons1(X1, X2))
proper(2nd(X))2nd(proper(X))from(ok(X))ok(from(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))active(2nd(cons1(X, cons(Y, Z))))mark(Y)
from(mark(X))mark(from(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
active(cons(X1, X2))cons(active(X1), X2)proper(cons(X1, X2))cons(proper(X1), proper(X2))
active(2nd(X))2nd(active(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
proper(from(X))from(proper(X))

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

top#(mark(X))top#(proper(X))

Problem 11: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(ok(X))top#(active(X))

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, ok, mark, proper, from, top, cons

Strategy


Polynomial Interpretation

Improved Usable rules

active(s(X))s(active(X))2nd(mark(X))mark(2nd(X))
s(ok(X))ok(s(X))cons(mark(X1), X2)mark(cons(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))from(ok(X))ok(from(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))active(from(X))from(active(X))
active(cons1(X1, X2))cons1(active(X1), X2)active(2nd(cons1(X, cons(Y, Z))))mark(Y)
from(mark(X))mark(from(X))active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
cons1(mark(X1), X2)mark(cons1(X1, X2))active(cons1(X1, X2))cons1(X1, active(X2))
active(cons(X1, X2))cons(active(X1), X2)s(mark(X))mark(s(X))
2nd(ok(X))ok(2nd(X))active(2nd(X))2nd(active(X))
active(from(X))mark(cons(X, from(s(X))))cons1(ok(X1), ok(X2))ok(cons1(X1, X2))

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

top#(ok(X))top#(active(X))

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

from#(mark(X))from#(X)from#(ok(X))from#(X)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

from#(mark(X))from#(X)from#(ok(X))from#(X)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

active#(from(X))active#(X)active#(s(X))active#(X)
active#(cons1(X1, X2))active#(X1)active#(cons1(X1, X2))active#(X2)
active#(2nd(X))active#(X)active#(cons(X1, X2))active#(X1)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

active#(s(X))active#(X)active#(from(X))active#(X)
active#(cons1(X1, X2))active#(X1)active#(cons1(X1, X2))active#(X2)
active#(cons(X1, X2))active#(X1)active#(2nd(X))active#(X)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

2nd#(ok(X))2nd#(X)2nd#(mark(X))2nd#(X)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

2nd#(ok(X))2nd#(X)2nd#(mark(X))2nd#(X)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons1#(mark(X1), X2)cons1#(X1, X2)cons1#(ok(X1), ok(X2))cons1#(X1, X2)
cons1#(X1, mark(X2))cons1#(X1, X2)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

cons1#(mark(X1), X2)cons1#(X1, X2)cons1#(ok(X1), ok(X2))cons1#(X1, X2)

Problem 10: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons1#(X1, mark(X2))cons1#(X1, X2)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, ok, mark, proper, from, top, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

cons1#(X1, mark(X2))cons1#(X1, X2)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

proper#(s(X))proper#(X)proper#(cons1(X1, X2))proper#(X2)
proper#(2nd(X))proper#(X)proper#(cons(X1, X2))proper#(X1)
proper#(cons(X1, X2))proper#(X2)proper#(cons1(X1, X2))proper#(X1)
proper#(from(X))proper#(X)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

proper#(cons1(X1, X2))proper#(X2)proper#(s(X))proper#(X)
proper#(cons(X1, X2))proper#(X1)proper#(2nd(X))proper#(X)
proper#(cons(X1, X2))proper#(X2)proper#(cons1(X1, X2))proper#(X1)
proper#(from(X))proper#(X)

Problem 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

s#(mark(X))s#(X)s#(ok(X))s#(X)

Rewrite Rules

active(2nd(cons1(X, cons(Y, Z))))mark(Y)active(2nd(cons(X, X1)))mark(2nd(cons1(X, X1)))
active(from(X))mark(cons(X, from(s(X))))active(2nd(X))2nd(active(X))
active(cons(X1, X2))cons(active(X1), X2)active(from(X))from(active(X))
active(s(X))s(active(X))active(cons1(X1, X2))cons1(active(X1), X2)
active(cons1(X1, X2))cons1(X1, active(X2))2nd(mark(X))mark(2nd(X))
cons(mark(X1), X2)mark(cons(X1, X2))from(mark(X))mark(from(X))
s(mark(X))mark(s(X))cons1(mark(X1), X2)mark(cons1(X1, X2))
cons1(X1, mark(X2))mark(cons1(X1, X2))proper(2nd(X))2nd(proper(X))
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
proper(s(X))s(proper(X))proper(cons1(X1, X2))cons1(proper(X1), proper(X2))
2nd(ok(X))ok(2nd(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))s(ok(X))ok(s(X))
cons1(ok(X1), ok(X2))ok(cons1(X1, X2))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: 2nd, cons1, s, active, mark, ok, from, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

s#(mark(X))s#(X)s#(ok(X))s#(X)