YES

The TRS could be proven terminating. The proof took 845 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (65ms).
 | – Problem 2 was processed with processor SubtermCriterion (4ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor PolynomialLinearRange4iUR (340ms).
 |    | – Problem 7 was processed with processor PolynomialLinearRange4iUR (363ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

proper#(f(X))f#(proper(X))proper#(f(X))proper#(X)
top#(mark(X))top#(proper(X))top#(ok(X))top#(active(X))
g#(ok(X))g#(X)top#(ok(X))active#(X)
active#(f(X))f#(active(X))active#(f(f(a)))f#(g(f(a)))
active#(f(f(a)))f#(a)proper#(g(X))g#(proper(X))
active#(f(f(a)))g#(f(a))proper#(g(X))proper#(X)
f#(mark(X))f#(X)active#(f(X))active#(X)
top#(mark(X))proper#(X)f#(ok(X))f#(X)

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


The following SCCs where found

f#(mark(X)) → f#(X)f#(ok(X)) → f#(X)

g#(ok(X)) → g#(X)

active#(f(X)) → active#(X)

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

proper#(f(X)) → proper#(X)proper#(g(X)) → proper#(X)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(ok(X))g#(X)

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(ok(X))g#(X)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(mark(X))f#(X)f#(ok(X))f#(X)

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(mark(X))f#(X)f#(ok(X))f#(X)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

active#(f(X))active#(X)

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

active#(f(X))active#(X)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

proper#(f(X))proper#(X)proper#(g(X))proper#(X)

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

proper#(f(X))proper#(X)proper#(g(X))proper#(X)

Problem 6: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, mark, ok, proper, top

Strategy


Polynomial Interpretation

Improved Usable rules

proper(g(X))g(proper(X))g(ok(X))ok(g(X))
proper(a)ok(a)active(f(f(a)))mark(f(g(f(a))))
proper(f(X))f(proper(X))f(mark(X))mark(f(X))
f(ok(X))ok(f(X))active(f(X))f(active(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 7: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(mark(X))mark(f(X))proper(f(X))f(proper(X))
proper(a)ok(a)proper(g(X))g(proper(X))
f(ok(X))ok(f(X))g(ok(X))ok(g(X))
top(mark(X))top(proper(X))top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: f, g, a, active, ok, mark, proper, top

Strategy


Polynomial Interpretation

Improved Usable rules

active(f(f(a)))mark(f(g(f(a))))active(f(X))f(active(X))
f(ok(X))ok(f(X))f(mark(X))mark(f(X))

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))