MAYBE
The TRS could not be proven terminating. The proof attempt took 19904 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| Problem 2 was processed with processor SubtermCriterion (0ms).
| | Problem 4 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | Problem 6 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | | Problem 8 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | | | Problem 10 was processed with processor PolynomialLinearRange4iUR (0ms).
| Problem 3 was processed with processor PolynomialLinearRange4iUR (0ms).
| | Problem 5 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | Problem 7 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | | Problem 9 was processed with processor PolynomialLinearRange4iUR (0ms).
| | | | | Problem 11 remains open; application of the following processors failed [DependencyGraph (2ms), PolynomialLinearRange4iUR (881ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (8485ms), DependencyGraph (1ms), ReductionPairSAT (7424ms), DependencyGraph (1ms), SizeChangePrinciple (47ms)].
The following open problems remain:
Open Dependency Pair Problem 11
Dependency Pairs
mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) | | active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
active#(f(a, b, X)) | → | f#(X, X, X) | | mark#(a) | → | active#(a) |
f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) | | mark#(f(X1, X2, X3)) | → | mark#(X3) |
active#(c) | → | mark#(a) | | mark#(c) | → | active#(c) |
f#(X1, X2, mark(X3)) | → | f#(X1, X2, X3) | | mark#(b) | → | active#(b) |
f#(mark(X1), X2, X3) | → | f#(X1, X2, X3) | | active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) | | mark#(f(X1, X2, X3)) | → | f#(X1, X2, mark(X3)) |
active#(c) | → | mark#(b) | | 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, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, 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#(X1, active(X2), X3) → f#(X1, X2, X3) |
f#(active(X1), X2, X3) → f#(X1, X2, X3) | f#(X1, X2, active(X3)) → f#(X1, X2, X3) |
mark#(b) → active#(b) | active#(f(a, b, X)) → mark#(f(X, X, X)) |
mark#(f(X1, X2, X3)) → active#(f(X1, X2, mark(X3))) | mark#(a) → active#(a) |
mark#(f(X1, X2, X3)) → mark#(X3) | active#(c) → mark#(b) |
mark#(c) → active#(c) | active#(c) → mark#(a) |
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#(X1, active(X2), X3) | → | f#(X1, X2, X3) |
f#(active(X1), X2, X3) | → | f#(X1, X2, X3) | | f#(X1, X2, active(X3)) | → | f#(X1, X2, X3) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, 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: PolynomialLinearRange4iUR
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, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): x + 1
- b: 0
- c: 0
- f(x,y,z): 0
- f#(x,y,z): z
- mark(x): 2x
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 6: PolynomialLinearRange4iUR
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) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 3
- b: 0
- c: 0
- f(x,y,z): 0
- f#(x,y,z): z
- 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 8: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) | | f#(X1, active(X2), X3) | → | f#(X1, X2, X3) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): x + 1
- b: 0
- c: 0
- f(x,y,z): 0
- f#(x,y,z): 2y
- mark(x): x
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, active(X2), X3) | → | f#(X1, X2, X3) |
Problem 10: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
f#(X1, mark(X2), X3) | → | f#(X1, X2, X3) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 0
- b: 0
- c: 0
- f(x,y,z): 0
- f#(x,y,z): z + 2y + x
- mark(x): 2x + 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, mark(X2), X3) | → | f#(X1, X2, X3) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(b) | → | active#(b) | | active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) | | mark#(a) | → | active#(a) |
mark#(f(X1, X2, X3)) | → | mark#(X3) | | active#(c) | → | mark#(b) |
mark#(c) | → | active#(c) | | active#(c) | → | mark#(a) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 2
- active(x): x
- active#(x): x
- b: 2
- c: 2
- f(x,y,z): z + 2
- mark(x): x
- mark#(x): x
Improved Usable rules
f(X1, X2, mark(X3)) | → | f(X1, X2, X3) | | active(f(a, b, X)) | → | mark(f(X, X, X)) |
f(mark(X1), X2, X3) | → | f(X1, X2, X3) | | f(active(X1), X2, X3) | → | f(X1, X2, X3) |
active(c) | → | mark(b) | | f(X1, X2, active(X3)) | → | f(X1, X2, X3) |
mark(c) | → | active(c) | | f(X1, active(X2), X3) | → | f(X1, X2, X3) |
active(c) | → | mark(a) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
f(X1, mark(X2), X3) | → | f(X1, X2, X3) | | mark(b) | → | active(b) |
mark(a) | → | active(a) |
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#(X3) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(b) | → | active#(b) | | mark#(a) | → | active#(a) |
mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) | | active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
active#(c) | → | mark#(b) | | active#(c) | → | mark#(a) |
mark#(c) | → | active#(c) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 2x
- active#(x): 2x
- b: 0
- c: 1
- f(x,y,z): 0
- mark(x): 1
- mark#(x): 3x
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#(c) | → | mark#(b) | | mark#(c) | → | active#(c) |
active#(c) | → | mark#(a) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(b) | → | active#(b) | | active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) | | mark#(a) | → | active#(a) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 1
- active#(x): 0
- b: 1
- c: 0
- f(x,y,z): 0
- mark(x): 2x
- 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:
Problem 9: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(a) | → | active#(a) | | mark#(f(X1, X2, X3)) | → | active#(f(X1, X2, mark(X3))) |
active#(f(a, b, X)) | → | mark#(f(X, X, X)) |
Rewrite Rules
active(f(a, b, X)) | → | mark(f(X, X, X)) | | active(c) | → | mark(a) |
active(c) | → | mark(b) | | mark(f(X1, X2, X3)) | → | active(f(X1, X2, mark(X3))) |
mark(a) | → | active(a) | | mark(b) | → | active(b) |
mark(c) | → | active(c) | | 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, c, a, active, mark
Strategy
Polynomial Interpretation
- a: 0
- active(x): 0
- active#(x): 2x
- b: 3
- c: 3
- f(x,y,z): 1
- mark(x): 1
- mark#(x): 2
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: