YES

The TRS could be proven terminating. The proof took 2363 ms.

The following DP Processors were used


Problem 1 was processed with processor PolynomialLinearRange4iUR (527ms).
 | – Problem 2 was processed with processor DependencyGraph (4ms).
 |    | – Problem 3 was processed with processor ForwardNarrowing (1ms).
 |    |    | – Problem 8 was processed with processor BackwardsNarrowing (1ms).
 |    |    |    | – Problem 11 was processed with processor BackwardsNarrowing (42ms).
 |    |    |    |    | – Problem 13 was processed with processor BackwardsNarrowing (1ms).
 |    |    |    |    |    | – Problem 14 was processed with processor BackwardsNarrowing (4ms).
 |    |    | – Problem 9 was processed with processor ForwardNarrowing (0ms).
 |    |    |    | – Problem 10 was processed with processor ForwardNarrowing (0ms).
 |    |    |    |    | – Problem 12 was processed with processor ForwardNarrowing (1ms).
 |    | – Problem 4 was processed with processor PolynomialLinearRange4iUR (138ms).
 |    |    | – Problem 5 was processed with processor DependencyGraph (1ms).
 |    |    |    | – Problem 6 was processed with processor PolynomialLinearRange4iUR (74ms).
 |    |    |    |    | – Problem 7 was processed with processor PolynomialLinearRange4iUR (28ms).

Problem 1: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

mark#(p(X))a__p#(mark(X))mark#(p(X))mark#(X)
mark#(cons(X1, X2))mark#(X1)mark#(f(X))a__f#(mark(X))
mark#(s(X))mark#(X)mark#(f(X))mark#(X)
a__f#(s(0))a__p#(s(0))a__p#(s(X))mark#(X)
a__f#(s(0))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


Polynomial Interpretation

Improved Usable rules

mark(cons(X1, X2))cons(mark(X1), X2)mark(f(X))a__f(mark(X))
a__p(X)p(X)mark(0)0
a__p(s(X))mark(X)mark(s(X))s(mark(X))
mark(p(X))a__p(mark(X))a__f(X)f(X)
a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))

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

mark#(f(X))a__f#(mark(X))mark#(f(X))mark#(X)
a__f#(s(0))a__p#(s(0))

Problem 2: DependencyGraph



Dependency Pair Problem

Dependency Pairs

mark#(p(X))mark#(X)mark#(p(X))a__p#(mark(X))
mark#(cons(X1, X2))mark#(X1)mark#(s(X))mark#(X)
a__p#(s(X))mark#(X)a__f#(s(0))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


The following SCCs where found

mark#(p(X)) → mark#(X)mark#(p(X)) → a__p#(mark(X))
mark#(cons(X1, X2)) → mark#(X1)mark#(s(X)) → mark#(X)
a__p#(s(X)) → mark#(X)

a__f#(s(0)) → a__f#(a__p(s(0)))

Problem 3: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(0))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


The right-hand side of the rule a__f#(s(0)) → a__f#(a__p(s(0))) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(mark(0)) 
a__f#(p(s(0))) 
Thus, the rule a__f#(s(0)) → a__f#(a__p(s(0))) is replaced by the following rules:
a__f#(s(0)) → a__f#(mark(0))a__f#(s(0)) → a__f#(p(s(0)))

Problem 8: BackwardsNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(mark(0)))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


The left-hand side of the rule a__f#(s(mark(0))) → a__f#(a__p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(mark(s(0))) 
a__f#(s(a__p(s(0)))) 
a__f#(s(mark(mark(0)))) 
Thus, the rule a__f#(s(mark(0))) → a__f#(a__p(s(0))) is replaced by the following rules:
a__f#(mark(s(0))) → a__f#(a__p(s(0)))a__f#(s(mark(mark(0)))) → a__f#(a__p(s(0)))
a__f#(s(a__p(s(0)))) → a__f#(a__p(s(0)))

Problem 11: BackwardsNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(mark(s(0)))a__f#(a__p(s(0)))a__f#(s(mark(mark(0))))a__f#(a__p(s(0)))
a__f#(s(a__p(s(0))))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


The left-hand side of the rule a__f#(s(mark(mark(0)))) → a__f#(a__p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(s(mark(a__p(s(0))))) 
a__f#(mark(s(mark(0)))) 
a__f#(s(mark(mark(mark(0))))) 
a__f#(s(a__p(s(mark(0))))) 
Thus, the rule a__f#(s(mark(mark(0)))) → a__f#(a__p(s(0))) is replaced by the following rules:
a__f#(s(mark(a__p(s(0))))) → a__f#(a__p(s(0)))a__f#(mark(s(mark(0)))) → a__f#(a__p(s(0)))
a__f#(s(mark(mark(mark(0))))) → a__f#(a__p(s(0)))a__f#(s(a__p(s(mark(0))))) → a__f#(a__p(s(0)))

Problem 13: BackwardsNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(mark(a__p(s(0)))))a__f#(a__p(s(0)))a__f#(mark(s(mark(0))))a__f#(a__p(s(0)))
a__f#(mark(s(0)))a__f#(a__p(s(0)))a__f#(s(mark(mark(mark(0)))))a__f#(a__p(s(0)))
a__f#(s(a__p(s(0))))a__f#(a__p(s(0)))a__f#(s(a__p(s(mark(0)))))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


The left-hand side of the rule a__f#(s(mark(a__p(s(0))))) → a__f#(a__p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(mark(s(a__p(s(0))))) 
a__f#(s(mark(a__p(s(mark(0)))))) 
a__f#(s(a__p(s(a__p(s(0)))))) 
Thus, the rule a__f#(s(mark(a__p(s(0))))) → a__f#(a__p(s(0))) is replaced by the following rules:
a__f#(mark(s(a__p(s(0))))) → a__f#(a__p(s(0)))a__f#(s(a__p(s(a__p(s(0)))))) → a__f#(a__p(s(0)))
a__f#(s(mark(a__p(s(mark(0)))))) → a__f#(a__p(s(0)))

Problem 14: BackwardsNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(mark(s(mark(0))))a__f#(a__p(s(0)))a__f#(mark(s(0)))a__f#(a__p(s(0)))
a__f#(s(mark(mark(mark(0)))))a__f#(a__p(s(0)))a__f#(s(a__p(s(mark(0)))))a__f#(a__p(s(0)))
a__f#(s(a__p(s(0))))a__f#(a__p(s(0)))a__f#(mark(s(a__p(s(0)))))a__f#(a__p(s(0)))
a__f#(s(a__p(s(a__p(s(0))))))a__f#(a__p(s(0)))a__f#(s(mark(a__p(s(mark(0))))))a__f#(a__p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


The left-hand side of the rule a__f#(mark(s(mark(0)))) → a__f#(a__p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(a__p(s(s(mark(0))))) 
a__f#(mark(s(a__p(s(0))))) 
a__f#(mark(s(mark(mark(0))))) 
a__f#(mark(mark(s(0)))) 
Thus, the rule a__f#(mark(s(mark(0)))) → a__f#(a__p(s(0))) is replaced by the following rules:
a__f#(a__p(s(s(mark(0))))) → a__f#(a__p(s(0)))a__f#(mark(s(mark(mark(0))))) → a__f#(a__p(s(0)))
a__f#(mark(s(a__p(s(0))))) → a__f#(a__p(s(0)))a__f#(mark(mark(s(0)))) → a__f#(a__p(s(0)))

Problem 9: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(0))a__f#(mark(0))a__f#(s(0))a__f#(p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


The right-hand side of the rule a__f#(s(0)) → a__f#(mark(0)) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
a__f#(0) 
Thus, the rule a__f#(s(0)) → a__f#(mark(0)) is replaced by the following rules:
a__f#(s(0)) → a__f#(0)

Problem 10: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(0))a__f#(0)a__f#(s(0))a__f#(p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


The right-hand side of the rule a__f#(s(0)) → a__f#(0) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
Thus, the rule a__f#(s(0)) → a__f#(0) is deleted.

Problem 12: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

a__f#(s(0))a__f#(p(s(0)))

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


The right-hand side of the rule a__f#(s(0)) → a__f#(p(s(0))) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
Thus, the rule a__f#(s(0)) → a__f#(p(s(0))) is deleted.

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

mark#(p(X))mark#(X)mark#(p(X))a__p#(mark(X))
mark#(cons(X1, X2))mark#(X1)mark#(s(X))mark#(X)
a__p#(s(X))mark#(X)

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


Polynomial Interpretation

Improved Usable rules

mark(cons(X1, X2))cons(mark(X1), X2)mark(f(X))a__f(mark(X))
a__p(X)p(X)mark(0)0
a__p(s(X))mark(X)mark(s(X))s(mark(X))
mark(p(X))a__p(mark(X))a__f(X)f(X)
a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))

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

mark#(s(X))mark#(X)a__p#(s(X))mark#(X)

Problem 5: DependencyGraph



Dependency Pair Problem

Dependency Pairs

mark#(p(X))a__p#(mark(X))mark#(p(X))mark#(X)
mark#(cons(X1, X2))mark#(X1)

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


The following SCCs where found

mark#(p(X)) → mark#(X)mark#(cons(X1, X2)) → mark#(X1)

Problem 6: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

mark#(p(X))mark#(X)mark#(cons(X1, X2))mark#(X1)

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons

Strategy


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:

mark#(p(X))mark#(X)

Problem 7: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

a__f(0)cons(0, f(s(0)))a__f(s(0))a__f(a__p(s(0)))
a__p(s(X))mark(X)mark(f(X))a__f(mark(X))
mark(p(X))a__p(mark(X))mark(0)0
mark(cons(X1, X2))cons(mark(X1), X2)mark(s(X))s(mark(X))
a__f(X)f(X)a__p(X)p(X)

Original Signature

Termination of terms over the following signature is verified: f, 0, a__p, s, p, mark, a__f, cons

Strategy


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:

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