YES
The TRS could be proven terminating. The proof took 355 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (38ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (226ms).
| | – Problem 3 was processed with processor DependencyGraph (20ms).
| | | – Problem 4 was processed with processor PolynomialLinearRange4iUR (19ms).
| | | | – Problem 5 was processed with processor DependencyGraph (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | add#(X1, X2) | | activate#(n__len(X)) | → | len#(X) |
| add#(s(X), Y) | → | activate#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(X1, X2) | | activate#(n__from(X)) | → | from#(X) |
| fst#(s(X), cons(Y, Z)) | → | activate#(X) | | fst#(s(X), cons(Y, Z)) | → | activate#(Z) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) |
| from(X) | → | cons(X, n__from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(n__add(activate(X), Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | fst(X1, X2) | → | n__fst(X1, X2) |
| from(X) | → | n__from(X) | | add(X1, X2) | → | n__add(X1, X2) |
| len(X) | → | n__len(X) | | activate(n__fst(X1, X2)) | → | fst(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
| activate(n__len(X)) | → | len(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, add, n__len, n__fst, fst, activate, 0, s, n__add, nil, cons
Strategy
The following SCCs where found
| activate#(n__add(X1, X2)) → add#(X1, X2) | activate#(n__len(X)) → len#(X) |
| add#(s(X), Y) → activate#(X) | len#(cons(X, Z)) → activate#(Z) |
| activate#(n__fst(X1, X2)) → fst#(X1, X2) | fst#(s(X), cons(Y, Z)) → activate#(X) |
| fst#(s(X), cons(Y, Z)) → activate#(Z) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | add#(X1, X2) | | activate#(n__len(X)) | → | len#(X) |
| add#(s(X), Y) | → | activate#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(X1, X2) | | fst#(s(X), cons(Y, Z)) | → | activate#(X) |
| fst#(s(X), cons(Y, Z)) | → | activate#(Z) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) |
| from(X) | → | cons(X, n__from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(n__add(activate(X), Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | fst(X1, X2) | → | n__fst(X1, X2) |
| from(X) | → | n__from(X) | | add(X1, X2) | → | n__add(X1, X2) |
| len(X) | → | n__len(X) | | activate(n__fst(X1, X2)) | → | fst(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
| activate(n__len(X)) | → | len(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, add, n__len, n__fst, fst, activate, 0, s, n__add, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): 2x + 2
- add(x,y): 0
- add#(x,y): 2x + 2
- cons(x,y): y + 1
- from(x): 0
- fst(x,y): 0
- fst#(x,y): 2y + 2x
- len(x): 0
- len#(x): 2x + 1
- n__add(x,y): 2x
- n__from(x): 0
- n__fst(x,y): y + 2x
- n__len(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:
| activate#(n__len(X)) | → | len#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(X1, X2) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | add#(X1, X2) | | add#(s(X), Y) | → | activate#(X) |
| fst#(s(X), cons(Y, Z)) | → | activate#(Z) | | fst#(s(X), cons(Y, Z)) | → | activate#(X) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) |
| from(X) | → | cons(X, n__from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(n__add(activate(X), Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | fst(X1, X2) | → | n__fst(X1, X2) |
| from(X) | → | n__from(X) | | add(X1, X2) | → | n__add(X1, X2) |
| len(X) | → | n__len(X) | | activate(n__fst(X1, X2)) | → | fst(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
| activate(n__len(X)) | → | len(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, add, n__len, n__fst, fst, activate, 0, s, n__add, nil, cons
Strategy
The following SCCs where found
| activate#(n__add(X1, X2)) → add#(X1, X2) | add#(s(X), Y) → activate#(X) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | add#(X1, X2) | | add#(s(X), Y) | → | activate#(X) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) |
| from(X) | → | cons(X, n__from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(n__add(activate(X), Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | fst(X1, X2) | → | n__fst(X1, X2) |
| from(X) | → | n__from(X) | | add(X1, X2) | → | n__add(X1, X2) |
| len(X) | → | n__len(X) | | activate(n__fst(X1, X2)) | → | fst(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
| activate(n__len(X)) | → | len(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, add, n__len, n__fst, fst, activate, 0, s, n__add, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): 2x + 2
- add(x,y): 0
- add#(x,y): x + 1
- cons(x,y): 0
- from(x): 0
- fst(x,y): 0
- len(x): 0
- n__add(x,y): x
- n__from(x): 0
- n__fst(x,y): 0
- n__len(x): 0
- nil: 0
- s(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:
| activate#(n__add(X1, X2)) | → | add#(X1, X2) |
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | activate#(X) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) |
| from(X) | → | cons(X, n__from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(n__add(activate(X), Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | fst(X1, X2) | → | n__fst(X1, X2) |
| from(X) | → | n__from(X) | | add(X1, X2) | → | n__add(X1, X2) |
| len(X) | → | n__len(X) | | activate(n__fst(X1, X2)) | → | fst(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__add(X1, X2)) | → | add(X1, X2) |
| activate(n__len(X)) | → | len(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, add, n__len, n__fst, fst, activate, 0, s, n__add, nil, cons
Strategy
There are no SCCs!