YES
The TRS could be proven terminating. The proof took 1383 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (719ms).
| – Problem 2 was processed with processor DependencyGraph (5ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (242ms).
| | | – Problem 4 was processed with processor PolynomialLinearRange4iUR (48ms).
| | | | – Problem 5 was processed with processor PolynomialLinearRange4iUR (53ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| a__from#(X) | → | mark#(X) | | mark#(first(X1, X2)) | → | mark#(X2) |
| mark#(from(X)) | → | a__from#(mark(X)) | | a__first#(s(X), cons(Y, Z)) | → | mark#(Y) |
| mark#(first(X1, X2)) | → | a__first#(mark(X1), mark(X2)) | | mark#(from(X)) | → | mark#(X) |
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(first(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__first(0, X) | → | nil | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(0) | → | 0 |
| mark(nil) | → | nil | | mark(s(X)) | → | s(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| a__from(X) | → | from(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, mark, from, a__first, first, a__from, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__first(x,y): y + x
- a__first#(x,y): 2y
- a__from(x): x + 1
- a__from#(x): 2x
- cons(x,y): x
- first(x,y): y + x
- from(x): x + 1
- mark(x): x
- mark#(x): 2x
- nil: 0
- s(x): x
Improved Usable rules
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| mark(s(X)) | → | s(mark(X)) | | a__from(X) | → | cons(mark(X), from(s(X))) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(nil) | → | nil |
| mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__first(0, X) | → | nil |
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 2: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(first(X1, X2)) | → | mark#(X2) | | a__from#(X) | → | mark#(X) |
| mark#(first(X1, X2)) | → | a__first#(mark(X1), mark(X2)) | | a__first#(s(X), cons(Y, Z)) | → | mark#(Y) |
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(first(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__first(0, X) | → | nil | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(0) | → | 0 |
| mark(nil) | → | nil | | mark(s(X)) | → | s(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| a__from(X) | → | from(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, mark, from, first, a__first, a__from, cons, nil
Strategy
The following SCCs where found
| mark#(first(X1, X2)) → mark#(X2) | mark#(first(X1, X2)) → a__first#(mark(X1), mark(X2)) |
| a__first#(s(X), cons(Y, Z)) → mark#(Y) | mark#(cons(X1, X2)) → mark#(X1) |
| mark#(first(X1, X2)) → mark#(X1) | mark#(s(X)) → mark#(X) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(first(X1, X2)) | → | mark#(X2) | | mark#(first(X1, X2)) | → | a__first#(mark(X1), mark(X2)) |
| a__first#(s(X), cons(Y, Z)) | → | mark#(Y) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(first(X1, X2)) | → | mark#(X1) | | mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__first(0, X) | → | nil | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(0) | → | 0 |
| mark(nil) | → | nil | | mark(s(X)) | → | s(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| a__from(X) | → | from(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, mark, from, first, a__first, a__from, cons, nil
Strategy
Polynomial Interpretation
- 0: 1
- a__first(x,y): 2y + x + 2
- a__first#(x,y): 2y + 1
- a__from(x): 2x
- cons(x,y): x
- first(x,y): 2y + x + 2
- from(x): 2x
- mark(x): x
- mark#(x): x
- nil: 3
- s(x): 2x
Improved Usable rules
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| mark(0) | → | 0 | | a__from(X) | → | from(X) |
| mark(s(X)) | → | s(mark(X)) | | a__from(X) | → | cons(mark(X), from(s(X))) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(nil) | → | nil |
| mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__first(0, X) | → | nil |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(first(X1, X2)) | → | mark#(X2) | | a__first#(s(X), cons(Y, Z)) | → | mark#(Y) |
| mark#(first(X1, X2)) | → | a__first#(mark(X1), mark(X2)) | | mark#(first(X1, X2)) | → | mark#(X1) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(s(X)) | → | mark#(X) |
Rewrite Rules
| a__first(0, X) | → | nil | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(0) | → | 0 |
| mark(nil) | → | nil | | mark(s(X)) | → | s(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| a__from(X) | → | from(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, mark, from, a__first, first, a__from, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__first(x,y): 0
- a__from(x): 0
- cons(x,y): x + 1
- first(x,y): 0
- from(x): 0
- mark(x): 0
- mark#(x): x
- nil: 0
- s(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:
| mark#(cons(X1, X2)) | → | mark#(X1) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| a__first(0, X) | → | nil | | a__first(s(X), cons(Y, Z)) | → | cons(mark(Y), first(X, Z)) |
| a__from(X) | → | cons(mark(X), from(s(X))) | | mark(first(X1, X2)) | → | a__first(mark(X1), mark(X2)) |
| mark(from(X)) | → | a__from(mark(X)) | | mark(0) | → | 0 |
| mark(nil) | → | nil | | mark(s(X)) | → | s(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | a__first(X1, X2) | → | first(X1, X2) |
| a__from(X) | → | from(X) |
Original Signature
Termination of terms over the following signature is verified: 0, s, mark, from, first, a__first, a__from, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- a__first(x,y): 0
- a__from(x): 0
- cons(x,y): 0
- first(x,y): 0
- from(x): 0
- mark(x): 0
- mark#(x): x
- nil: 0
- s(x): x + 2
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: