YES
The TRS could be proven terminating. The proof took 685 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (37ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (317ms).
| | – Problem 5 was processed with processor DependencyGraph (15ms).
| | | – Problem 6 was processed with processor PolynomialLinearRange4iUR (26ms).
| | | | – Problem 7 was processed with processor PolynomialLinearRange4iUR (58ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor DependencyGraph (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__from#(X) | → | mark#(X) | | a__length#(cons(X, Y)) | → | a__length1#(Y) |
| a__length1#(X) | → | a__length#(X) | | mark#(from(X)) | → | a__from#(mark(X)) |
| mark#(from(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) | | mark#(length1(X)) | → | a__length1#(X) |
| mark#(length(X)) | → | a__length#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, from, length1, a__length1, a__from, cons, nil
Strategy
The following SCCs where found
| a__from#(X) → mark#(X) | mark#(from(X)) → a__from#(mark(X)) |
| mark#(from(X)) → mark#(X) | mark#(cons(X1, X2)) → mark#(X1) |
| mark#(s(X)) → mark#(X) |
| a__length#(cons(X, Y)) → a__length1#(Y) | a__length1#(X) → a__length#(X) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| a__from#(X) | → | mark#(X) | | mark#(from(X)) | → | a__from#(mark(X)) |
| mark#(from(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, from, length1, a__length1, a__from, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- a__from(x): 2x + 2
- a__from#(x): 2x
- a__length(x): 0
- a__length1(x): 0
- cons(x,y): 2x
- from(x): 2x + 2
- length(x): 0
- length1(x): 0
- mark(x): x
- mark#(x): x
- nil: 0
- s(x): 2x
Improved Usable rules
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| a__from(X) | → | from(X) | | mark(length1(X)) | → | a__length1(X) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length1(X) | → | a__length(X) |
| mark(length(X)) | → | a__length(X) | | a__length(X) | → | length(X) |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | mark(s(X)) | → | s(mark(X)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(nil) | → | nil |
| a__length1(X) | → | length1(X) | | a__length(nil) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(from(X)) | → | a__from#(mark(X)) | | mark#(from(X)) | → | mark#(X) |
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__from#(X) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, length1, from, a__length1, a__from, nil, cons
Strategy
The following SCCs where found
| mark#(cons(X1, X2)) → mark#(X1) | mark#(s(X)) → mark#(X) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, length1, from, a__length1, a__from, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__from(x): 0
- a__length(x): 0
- a__length1(x): 0
- cons(x,y): 2x
- from(x): 0
- length(x): 0
- length1(x): 0
- mark(x): 0
- mark#(x): 2x
- nil: 0
- s(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:
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(cons(X1, X2)) | → | mark#(X1) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, from, length1, a__length1, a__from, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- a__from(x): 0
- a__length(x): 0
- a__length1(x): 0
- cons(x,y): x + 2
- from(x): 0
- length(x): 0
- length1(x): 0
- mark(x): 0
- mark#(x): x + 1
- nil: 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) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| a__length#(cons(X, Y)) | → | a__length1#(Y) | | a__length1#(X) | → | a__length#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, from, length1, a__length1, a__from, cons, nil
Strategy
Projection
The following projection was used:
- π (a__length#): 1
- π (a__length1#): 1
Thus, the following dependency pairs are removed:
| a__length#(cons(X, Y)) | → | a__length1#(Y) |
Problem 4: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__length1#(X) | → | a__length#(X) |
Rewrite Rules
| a__from(X) | → | cons(mark(X), from(s(X))) | | a__length(nil) | → | 0 |
| a__length(cons(X, Y)) | → | s(a__length1(Y)) | | a__length1(X) | → | a__length(X) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(length(X)) | → | a__length(X) |
| mark(length1(X)) | → | a__length1(X) | | mark(cons(X1, X2)) | → | cons(mark(X1), X2) |
| mark(s(X)) | → | s(mark(X)) | | mark(nil) | → | nil |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| a__length(X) | → | length(X) | | a__length1(X) | → | length1(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, a__length, length, mark, length1, from, a__length1, a__from, nil, cons
Strategy
There are no SCCs!