YES
The TRS could be proven terminating. The proof took 1083 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (107ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor SubtermCriterion (1ms).
| | | – Problem 5 was processed with processor PolynomialLinearRange4iUR (51ms).
| | | | – Problem 7 was processed with processor PolynomialLinearRange4iUR (12ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (524ms).
| | – Problem 6 was processed with processor PolynomialLinearRange4iUR (260ms).
| | | – Problem 8 was processed with processor DependencyGraph (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2, X3)) | → | active#(f(X1, mark(X2), X3)) | | mark#(a) | → | active#(a) |
| f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) | | active#(f(a, X, X)) | → | mark#(f(X, b, b)) |
| active#(f(a, X, X)) | → | f#(X, b, b) | | f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) |
| mark#(b) | → | active#(b) | | active#(b) | → | mark#(a) |
| f#(mark(X1), X2, X3) | → | f#(X1, X2, X3) | | f#(active(X1), X2, X3) | → | f#(X1, X2, X3) |
| f#(X1, active(X2), X3) | → | f#(X1, X2, X3) | | mark#(f(X1, X2, X3)) | → | f#(X1, mark(X2), X3) |
| mark#(f(X1, X2, X3)) | → | mark#(X2) | | f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
The following SCCs where found
| f#(X1, X2, mark(X3)) → f#(X1, X2, X3) | f#(mark(X1), X2, X3) → f#(X1, X2, X3) |
| f#(X1, mark(X2), X3) → f#(X1, X2, X3) | f#(active(X1), X2, X3) → f#(X1, X2, X3) |
| f#(X1, active(X2), X3) → f#(X1, X2, X3) | f#(X1, X2, active(X3)) → f#(X1, X2, X3) |
| mark#(f(X1, X2, X3)) → active#(f(X1, mark(X2), X3)) | active#(f(a, X, X)) → mark#(f(X, b, b)) |
| mark#(f(X1, X2, X3)) → mark#(X2) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) | | f#(mark(X1), X2, X3) | → | f#(X1, X2, X3) |
| f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) | | f#(active(X1), X2, X3) | → | f#(X1, X2, X3) |
| f#(X1, active(X2), X3) | → | f#(X1, X2, X3) | | f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(mark(X1), X2, X3) | → | f#(X1, X2, X3) | | f#(active(X1), X2, X3) | → | f#(X1, X2, X3) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) | | f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) |
| f#(X1, active(X2), X3) | → | f#(X1, X2, X3) | | f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) | | f#(X1, active(X2), X3) | → | f#(X1, X2, X3) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) | | f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): x
- b: 0
- f(x,y,z): 0
- f#(x,y,z): 2z
- mark(x): x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 2x + 1
- b: 0
- f(x,y,z): 0
- f#(x,y,z): 2z + y + x
- mark(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:
| f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2, X3)) | → | active#(f(X1, mark(X2), X3)) | | active#(f(a, X, X)) | → | mark#(f(X, b, b)) |
| mark#(f(X1, X2, X3)) | → | mark#(X2) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): x + 1
- active#(x): 2
- b: 0
- f(x,y,z): y + 2
- mark(x): x
- mark#(x): x
Improved Usable rules
| f(X1, X2, mark(X3)) | → | f(X1, X2, X3) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, mark(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(f(X1, X2, X3)) | → | mark#(X2) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2, X3)) | → | active#(f(X1, mark(X2), X3)) | | active#(f(a, X, X)) | → | mark#(f(X, b, b)) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
Polynomial Interpretation
- a: 1
- active(x): x + 1
- active#(x): 2x
- b: 0
- f(x,y,z): 2z + x
- mark(x): 3x
- mark#(x): 2x
Improved Usable rules
| f(X1, X2, mark(X3)) | → | f(X1, X2, X3) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, mark(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| active#(f(a, X, X)) | → | mark#(f(X, b, b)) |
Problem 8: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2, X3)) | → | active#(f(X1, mark(X2), X3)) |
Rewrite Rules
| active(f(a, X, X)) | → | mark(f(X, b, b)) | | active(b) | → | mark(a) |
| mark(f(X1, X2, X3)) | → | active(f(X1, mark(X2), X3)) | | mark(a) | → | active(a) |
| mark(b) | → | active(b) | | f(mark(X1), X2, X3) | → | f(X1, X2, X3) |
| f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | f(X1, X2, mark(X3)) | → | f(X1, X2, X3) |
| f(active(X1), X2, X3) | → | f(X1, X2, X3) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
| f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, active, mark
Strategy
There are no SCCs!