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
- 0: 0
- a__f(x): x + 2
- a__f#(x): 2
- a__p(x): 2x
- a__p#(x): 2x + 1
- cons(x,y): x
- f(x): x + 2
- mark(x): x
- mark#(x): x + 1
- p(x): 2x
- s(x): 2x
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 Terms | Irrelevant 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 Terms | Irrelevant 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 Terms | Irrelevant 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 Terms | Irrelevant 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 Terms | Irrelevant 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 Terms | Irrelevant Terms |
|---|
| a__f#(0) | |
Thus, the rule a__f
#(s(0)) → a__f
#(mark(0)) is replaced by the following rules:
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 Terms | Irrelevant 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 Terms | Irrelevant 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
- 0: 0
- a__f(x): 0
- a__p(x): 2x
- a__p#(x): 2x
- cons(x,y): x
- f(x): 0
- mark(x): x
- mark#(x): x
- p(x): 2x
- s(x): x + 1
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
- 0: 0
- a__f(x): 0
- a__p(x): 0
- cons(x,y): x
- f(x): 0
- mark(x): 0
- mark#(x): x
- p(x): x + 1
- s(x): 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
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
- 0: 0
- a__f(x): 0
- a__p(x): 0
- cons(x,y): x + 1
- f(x): 0
- mark(x): 0
- mark#(x): x + 1
- p(x): 0
- s(x): 0
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) |