YES
The TRS could be proven terminating. The proof took 15596 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (429ms).
| Problem 2 was processed with processor SubtermCriterion (0ms).
| Problem 3 was processed with processor SubtermCriterion (33ms).
| Problem 4 was processed with processor SubtermCriterion (0ms).
| Problem 5 was processed with processor PolynomialLinearRange4iUR (7018ms).
| | Problem 9 was processed with processor PolynomialLinearRange4iUR (3542ms).
| | | Problem 10 was processed with processor PolynomialLinearRange4iUR (1966ms).
| | | | Problem 11 was processed with processor PolynomialLinearRange4iUR (2036ms).
| | | | | Problem 12 was processed with processor PolynomialLinearRange4iUR (67ms).
| | | | | | Problem 13 was processed with processor PolynomialLinearRange4iUR (89ms).
| | | | | | | Problem 14 was processed with processor PolynomialLinearRange4iUR (67ms).
| | | | | | | | Problem 15 was processed with processor PolynomialLinearRange4iUR (126ms).
| | | | | | | | | Problem 16 was processed with processor PolynomialLinearRange4iUR (44ms).
| | | | | | | | | | Problem 17 was processed with processor DependencyGraph (0ms).
| Problem 6 was processed with processor SubtermCriterion (1ms).
| | Problem 8 was processed with processor SubtermCriterion (1ms).
| Problem 7 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
mark#(cons(X1, X2)) | → | active#(cons(mark(X1), X2)) | | cons#(mark(X1), X2) | → | cons#(X1, X2) |
active#(length(nil)) | → | mark#(0) | | mark#(length1(X)) | → | active#(length1(X)) |
from#(mark(X)) | → | from#(X) | | mark#(s(X)) | → | s#(mark(X)) |
mark#(nil) | → | active#(nil) | | length#(active(X)) | → | length#(X) |
mark#(length(X)) | → | active#(length(X)) | | active#(from(X)) | → | cons#(X, from(s(X))) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(s(X)) | → | mark#(X) |
cons#(X1, mark(X2)) | → | cons#(X1, X2) | | mark#(cons(X1, X2)) | → | cons#(mark(X1), X2) |
length#(mark(X)) | → | length#(X) | | mark#(length(X)) | → | length#(X) |
mark#(0) | → | active#(0) | | mark#(s(X)) | → | active#(s(mark(X))) |
mark#(from(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
cons#(active(X1), X2) | → | cons#(X1, X2) | | active#(length(cons(X, Y))) | → | length1#(Y) |
active#(from(X)) | → | mark#(cons(X, from(s(X)))) | | length1#(mark(X)) | → | length1#(X) |
mark#(length1(X)) | → | length1#(X) | | active#(from(X)) | → | s#(X) |
s#(mark(X)) | → | s#(X) | | active#(length(cons(X, Y))) | → | s#(length1(Y)) |
cons#(X1, active(X2)) | → | cons#(X1, X2) | | mark#(from(X)) | → | from#(mark(X)) |
active#(length1(X)) | → | mark#(length(X)) | | mark#(from(X)) | → | active#(from(mark(X))) |
from#(active(X)) | → | from#(X) | | s#(active(X)) | → | s#(X) |
active#(length1(X)) | → | length#(X) | | active#(from(X)) | → | from#(s(X)) |
length1#(active(X)) | → | length1#(X) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
The following SCCs where found
length#(mark(X)) → length#(X) | length#(active(X)) → length#(X) |
from#(active(X)) → from#(X) | from#(mark(X)) → from#(X) |
length1#(mark(X)) → length1#(X) | length1#(active(X)) → length1#(X) |
cons#(X1, active(X2)) → cons#(X1, X2) | cons#(mark(X1), X2) → cons#(X1, X2) |
cons#(X1, mark(X2)) → cons#(X1, X2) | cons#(active(X1), X2) → cons#(X1, X2) |
s#(mark(X)) → s#(X) | s#(active(X)) → s#(X) |
mark#(cons(X1, X2)) → active#(cons(mark(X1), X2)) | mark#(s(X)) → active#(s(mark(X))) |
active#(length1(X)) → mark#(length(X)) | mark#(from(X)) → active#(from(mark(X))) |
mark#(from(X)) → mark#(X) | mark#(length(X)) → active#(length(X)) |
active#(length(cons(X, Y))) → mark#(s(length1(Y))) | mark#(cons(X1, X2)) → mark#(X1) |
mark#(s(X)) → mark#(X) | mark#(length1(X)) → active#(length1(X)) |
active#(from(X)) → mark#(cons(X, from(s(X)))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
length1#(mark(X)) | → | length1#(X) | | length1#(active(X)) | → | length1#(X) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
length1#(mark(X)) | → | length1#(X) | | length1#(active(X)) | → | length1#(X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
s#(mark(X)) | → | s#(X) | | s#(active(X)) | → | s#(X) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
s#(mark(X)) | → | s#(X) | | s#(active(X)) | → | s#(X) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
from#(active(X)) | → | from#(X) | | from#(mark(X)) | → | from#(X) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
from#(active(X)) | → | from#(X) | | from#(mark(X)) | → | from#(X) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(cons(X1, X2)) | → | active#(cons(mark(X1), X2)) | | mark#(s(X)) | → | active#(s(mark(X))) |
active#(length1(X)) | → | mark#(length(X)) | | mark#(from(X)) | → | active#(from(mark(X))) |
mark#(from(X)) | → | mark#(X) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(cons(X1, X2)) | → | mark#(X1) |
mark#(s(X)) | → | mark#(X) | | mark#(length1(X)) | → | active#(length1(X)) |
active#(from(X)) | → | mark#(cons(X, from(s(X)))) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Polynomial Interpretation
- 0: 2
- active(x): 2
- active#(x): x
- cons(x,y): 0
- from(x): 2
- length(x): 2
- length1(x): 2
- mark(x): 3
- mark#(x): 2
- nil: 1
- s(x): 2
Improved Usable rules
length1(mark(X)) | → | length1(X) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | from(mark(X)) | → | from(X) |
length1(active(X)) | → | length1(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
length(active(X)) | → | length(X) | | s(mark(X)) | → | s(X) |
length(mark(X)) | → | length(X) | | s(active(X)) | → | s(X) |
cons(X1, active(X2)) | → | cons(X1, X2) | | from(active(X)) | → | from(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(cons(X1, X2)) | → | active#(cons(mark(X1), X2)) |
Problem 9: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(s(X)) | → | active#(s(mark(X))) | | mark#(from(X)) | → | active#(from(mark(X))) |
active#(length1(X)) | → | mark#(length(X)) | | mark#(length(X)) | → | active#(length(X)) |
mark#(from(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(s(X)) | → | mark#(X) |
active#(from(X)) | → | mark#(cons(X, from(s(X)))) | | mark#(length1(X)) | → | active#(length1(X)) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
Polynomial Interpretation
- 0: 3
- active(x): 3
- active#(x): 2x
- cons(x,y): 1
- from(x): 1
- length(x): 1
- length1(x): 1
- mark(x): 0
- mark#(x): 2
- nil: 0
- s(x): 0
Improved Usable rules
length1(mark(X)) | → | length1(X) | | from(mark(X)) | → | from(X) |
length1(active(X)) | → | length1(X) | | s(mark(X)) | → | s(X) |
length(active(X)) | → | length(X) | | length(mark(X)) | → | length(X) |
s(active(X)) | → | s(X) | | from(active(X)) | → | from(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(s(X)) | → | active#(s(mark(X))) |
Problem 10: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
active#(length1(X)) | → | mark#(length(X)) | | mark#(from(X)) | → | active#(from(mark(X))) |
mark#(from(X)) | → | mark#(X) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(cons(X1, X2)) | → | mark#(X1) |
mark#(s(X)) | → | mark#(X) | | mark#(length1(X)) | → | active#(length1(X)) |
active#(from(X)) | → | mark#(cons(X, from(s(X)))) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- active(x): x
- active#(x): 2x
- cons(x,y): x
- from(x): x + 1
- length(x): 0
- length1(x): 0
- mark(x): x
- mark#(x): 2x
- nil: 0
- s(x): x
Improved Usable rules
cons(active(X1), X2) | → | cons(X1, X2) | | from(mark(X)) | → | from(X) |
mark(length(X)) | → | active(length(X)) | | length1(active(X)) | → | length1(X) |
mark(s(X)) | → | active(s(mark(X))) | | active(length1(X)) | → | mark(length(X)) |
length(mark(X)) | → | length(X) | | active(length(nil)) | → | mark(0) |
length1(mark(X)) | → | length1(X) | | cons(X1, mark(X2)) | → | cons(X1, X2) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | cons(mark(X1), X2) | → | cons(X1, X2) |
mark(length1(X)) | → | active(length1(X)) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(from(X)) | → | active(from(mark(X))) | | length(active(X)) | → | length(X) |
s(mark(X)) | → | s(X) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
s(active(X)) | → | s(X) | | from(active(X)) | → | from(X) |
cons(X1, active(X2)) | → | cons(X1, X2) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(from(X)) | → | mark#(X) | | active#(from(X)) | → | mark#(cons(X, from(s(X)))) |
Problem 11: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(from(X)) | → | active#(from(mark(X))) | | active#(length1(X)) | → | mark#(length(X)) |
mark#(length(X)) | → | active#(length(X)) | | mark#(cons(X1, X2)) | → | mark#(X1) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(s(X)) | → | mark#(X) |
mark#(length1(X)) | → | active#(length1(X)) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
Polynomial Interpretation
- 0: 1
- active(x): 2x
- active#(x): 1
- cons(x,y): x
- from(x): 1
- length(x): 0
- length1(x): 0
- mark(x): x
- mark#(x): 2x + 1
- nil: 0
- s(x): 2x
Improved Usable rules
length1(mark(X)) | → | length1(X) | | length1(active(X)) | → | length1(X) |
length(active(X)) | → | length(X) | | s(mark(X)) | → | s(X) |
length(mark(X)) | → | length(X) | | s(active(X)) | → | s(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(from(X)) | → | active#(from(mark(X))) |
Problem 12: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
active#(length1(X)) | → | mark#(length(X)) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(cons(X1, X2)) | → | mark#(X1) |
mark#(s(X)) | → | mark#(X) | | mark#(length1(X)) | → | active#(length1(X)) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- active(x): x
- active#(x): 2
- cons(x,y): x + 1
- from(x): 0
- length(x): 1
- length1(x): 1
- mark(x): x
- mark#(x): 2x
- nil: 0
- s(x): x
Improved Usable rules
length1(mark(X)) | → | length1(X) | | length1(active(X)) | → | length1(X) |
length(active(X)) | → | length(X) | | s(mark(X)) | → | s(X) |
length(mark(X)) | → | length(X) | | s(active(X)) | → | s(X) |
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) |
Problem 13: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
active#(length1(X)) | → | mark#(length(X)) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(s(X)) | → | mark#(X) |
mark#(length1(X)) | → | active#(length1(X)) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- active(x): x
- active#(x): 2x
- cons(x,y): y + 2x + 1
- from(x): 0
- length(x): x
- length1(x): x
- mark(x): x
- mark#(x): 2x
- nil: 0
- s(x): x + 1
Improved Usable rules
length1(mark(X)) | → | length1(X) | | length1(active(X)) | → | length1(X) |
length(active(X)) | → | length(X) | | s(mark(X)) | → | s(X) |
length(mark(X)) | → | length(X) | | s(active(X)) | → | s(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 14: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
active#(length1(X)) | → | mark#(length(X)) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) | | mark#(length1(X)) | → | active#(length1(X)) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- active(x): 0
- active#(x): 1
- cons(x,y): 3y + 3x + 2
- from(x): 0
- length(x): 0
- length1(x): x + 1
- mark(x): 0
- mark#(x): 2x + 1
- nil: 0
- s(x): 0
Improved Usable rules
length(active(X)) | → | length(X) | | s(mark(X)) | → | s(X) |
length(mark(X)) | → | length(X) | | s(active(X)) | → | s(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(length1(X)) | → | active#(length1(X)) |
Problem 15: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
active#(length1(X)) | → | mark#(length(X)) | | mark#(length(X)) | → | active#(length(X)) |
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- active(x): 0
- active#(x): 2x
- cons(x,y): 3y + x + 3
- from(x): 0
- length(x): 0
- length1(x): 1
- mark(x): 3x
- mark#(x): 0
- nil: 0
- s(x): 2x + 1
Improved Usable rules
length(active(X)) | → | length(X) | | length(mark(X)) | → | length(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
active#(length1(X)) | → | mark#(length(X)) |
Problem 16: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
mark#(length(X)) | → | active#(length(X)) | | active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- active(x): 0
- active#(x): 0
- cons(x,y): 1
- from(x): 0
- length(x): x + 1
- length1(x): x
- mark(x): 2
- mark#(x): x
- nil: 0
- s(x): 0
Improved Usable rules
s(mark(X)) | → | s(X) | | s(active(X)) | → | s(X) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
mark#(length(X)) | → | active#(length(X)) |
Problem 17: DependencyGraph
Dependency Pair Problem
Dependency Pairs
active#(length(cons(X, Y))) | → | mark#(s(length1(Y))) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
There are no SCCs!
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
cons#(X1, active(X2)) | → | cons#(X1, X2) | | cons#(mark(X1), X2) | → | cons#(X1, X2) |
cons#(X1, mark(X2)) | → | cons#(X1, X2) | | cons#(active(X1), X2) | → | cons#(X1, X2) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
cons#(mark(X1), X2) | → | cons#(X1, X2) | | cons#(active(X1), X2) | → | cons#(X1, X2) |
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
cons#(X1, active(X2)) | → | cons#(X1, X2) | | cons#(X1, mark(X2)) | → | cons#(X1, X2) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, length1, from, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
cons#(X1, active(X2)) | → | cons#(X1, X2) | | cons#(X1, mark(X2)) | → | cons#(X1, X2) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
length#(mark(X)) | → | length#(X) | | length#(active(X)) | → | length#(X) |
Rewrite Rules
active(from(X)) | → | mark(cons(X, from(s(X)))) | | active(length(nil)) | → | mark(0) |
active(length(cons(X, Y))) | → | mark(s(length1(Y))) | | active(length1(X)) | → | mark(length(X)) |
mark(from(X)) | → | active(from(mark(X))) | | mark(cons(X1, X2)) | → | active(cons(mark(X1), X2)) |
mark(s(X)) | → | active(s(mark(X))) | | mark(length(X)) | → | active(length(X)) |
mark(nil) | → | active(nil) | | mark(0) | → | active(0) |
mark(length1(X)) | → | active(length1(X)) | | from(mark(X)) | → | from(X) |
from(active(X)) | → | from(X) | | cons(mark(X1), X2) | → | cons(X1, X2) |
cons(X1, mark(X2)) | → | cons(X1, X2) | | cons(active(X1), X2) | → | cons(X1, X2) |
cons(X1, active(X2)) | → | cons(X1, X2) | | s(mark(X)) | → | s(X) |
s(active(X)) | → | s(X) | | length(mark(X)) | → | length(X) |
length(active(X)) | → | length(X) | | length1(mark(X)) | → | length1(X) |
length1(active(X)) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, length, active, mark, from, length1, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
length#(mark(X)) | → | length#(X) | | length#(active(X)) | → | length#(X) |