YES
The TRS could be proven terminating. The proof took 3419 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (185ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (1444ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (753ms).
| | | – Problem 4 was processed with processor DependencyGraph (9ms).
| | | | – Problem 5 was processed with processor PolynomialLinearRange4iUR (602ms).
| | | | | – Problem 6 was processed with processor DependencyGraph (4ms).
| | | | | | – Problem 7 was processed with processor PolynomialLinearRange4iUR (258ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | activate#(X2) | | add#(s(X), Y) | → | activate#(X) |
| activate#(n__fst(X1, X2)) | → | activate#(X2) | | activate#(n__from(X)) | → | activate#(X) |
| activate#(n__len(X)) | → | len#(activate(X)) | | activate#(n__s(X)) | → | s#(X) |
| add#(s(X), Y) | → | s#(n__add(activate(X), Y)) | | activate#(n__add(X1, X2)) | → | activate#(X1) |
| activate#(n__len(X)) | → | activate#(X) | | len#(cons(X, Z)) | → | s#(n__len(activate(Z))) |
| activate#(n__from(X)) | → | from#(activate(X)) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) | | activate#(n__fst(X1, X2)) | → | activate#(X1) |
| activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(X2)) | | 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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, cons, nil
Strategy
The following SCCs where found
| activate#(n__add(X1, X2)) → activate#(X2) | activate#(n__fst(X1, X2)) → activate#(X2) |
| add#(s(X), Y) → activate#(X) | activate#(n__from(X)) → activate#(X) |
| activate#(n__len(X)) → len#(activate(X)) | activate#(n__add(X1, X2)) → activate#(X1) |
| activate#(n__len(X)) → activate#(X) | len#(cons(X, Z)) → activate#(Z) |
| activate#(n__fst(X1, X2)) → fst#(activate(X1), activate(X2)) | activate#(n__fst(X1, X2)) → activate#(X1) |
| activate#(n__add(X1, X2)) → add#(activate(X1), activate(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
| add#(s(X), Y) | → | activate#(X) | | activate#(n__fst(X1, X2)) | → | activate#(X2) |
| activate#(n__add(X1, X2)) | → | activate#(X2) | | activate#(n__from(X)) | → | activate#(X) |
| activate#(n__len(X)) | → | len#(activate(X)) | | activate#(n__add(X1, X2)) | → | activate#(X1) |
| activate#(n__len(X)) | → | activate#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) | | activate#(n__fst(X1, X2)) | → | activate#(X1) |
| activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(X2)) | | 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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): 2x
- add(x,y): y + x
- add#(x,y): 2x
- cons(x,y): y
- from(x): 2x + 1
- fst(x,y): y + x
- fst#(x,y): 2y + 2x
- len(x): x
- len#(x): 2x
- n__add(x,y): y + x
- n__from(x): 2x + 1
- n__fst(x,y): y + x
- n__len(x): x
- n__s(x): x
- nil: 0
- s(x): x
Improved Usable rules
| from(X) | → | cons(X, n__from(n__s(X))) | | len(X) | → | n__len(X) |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__s(X)) | → | s(X) |
| add(0, X) | → | X | | add(X1, X2) | → | n__add(X1, X2) |
| len(nil) | → | 0 | | fst(0, Z) | → | nil |
| fst(X1, X2) | → | n__fst(X1, X2) | | s(X) | → | n__s(X) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) | | add(s(X), Y) | → | s(n__add(activate(X), Y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__from(X)) | → | activate#(X) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | activate#(X1) | | activate#(n__len(X)) | → | activate#(X) |
| activate#(n__add(X1, X2)) | → | activate#(X2) | | activate#(n__fst(X1, X2)) | → | activate#(X2) |
| add#(s(X), Y) | → | activate#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
| activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) | | activate#(n__fst(X1, X2)) | → | activate#(X1) |
| activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(X2)) | | activate#(n__len(X)) | → | len#(activate(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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): x
- add(x,y): 2y + 2x
- add#(x,y): 2y + x
- cons(x,y): y
- from(x): 0
- fst(x,y): 2y + 2x
- fst#(x,y): 2y + x
- len(x): 3x + 2
- len#(x): 2x + 2
- n__add(x,y): 2y + 2x
- n__from(x): 0
- n__fst(x,y): 2y + 2x
- n__len(x): 3x + 2
- n__s(x): x
- nil: 0
- s(x): x
Improved Usable rules
| from(X) | → | cons(X, n__from(n__s(X))) | | len(X) | → | n__len(X) |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__s(X)) | → | s(X) |
| add(0, X) | → | X | | add(X1, X2) | → | n__add(X1, X2) |
| len(nil) | → | 0 | | fst(0, Z) | → | nil |
| fst(X1, X2) | → | n__fst(X1, X2) | | s(X) | → | n__s(X) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) | | add(s(X), Y) | → | s(n__add(activate(X), Y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__len(X)) | → | activate#(X) | | len#(cons(X, Z)) | → | activate#(Z) |
Problem 4: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | activate#(X1) | | add#(s(X), Y) | → | activate#(X) |
| activate#(n__fst(X1, X2)) | → | activate#(X2) | | activate#(n__add(X1, X2)) | → | activate#(X2) |
| activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) | | activate#(n__fst(X1, X2)) | → | activate#(X1) |
| activate#(n__len(X)) | → | len#(activate(X)) | | activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(X2)) |
| 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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, cons, nil
Strategy
The following SCCs where found
| activate#(n__add(X1, X2)) → activate#(X1) | activate#(n__fst(X1, X2)) → activate#(X2) |
| add#(s(X), Y) → activate#(X) | activate#(n__add(X1, X2)) → activate#(X2) |
| activate#(n__fst(X1, X2)) → fst#(activate(X1), activate(X2)) | activate#(n__fst(X1, X2)) → activate#(X1) |
| activate#(n__add(X1, X2)) → add#(activate(X1), activate(X2)) | fst#(s(X), cons(Y, Z)) → activate#(Z) |
| fst#(s(X), cons(Y, Z)) → activate#(X) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__add(X1, X2)) | → | activate#(X1) | | activate#(n__fst(X1, X2)) | → | activate#(X2) |
| add#(s(X), Y) | → | activate#(X) | | activate#(n__add(X1, X2)) | → | activate#(X2) |
| activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) | | activate#(n__fst(X1, X2)) | → | activate#(X1) |
| activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(X2)) | | 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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): x
- add(x,y): y + 2x + 2
- add#(x,y): y + 2x + 2
- cons(x,y): y
- from(x): 2x + 1
- fst(x,y): y + x
- fst#(x,y): y + x
- len(x): 2x + 3
- n__add(x,y): y + 2x + 2
- n__from(x): 2x + 1
- n__fst(x,y): y + x
- n__len(x): 2x + 3
- n__s(x): x
- nil: 0
- s(x): x
Improved Usable rules
| from(X) | → | cons(X, n__from(n__s(X))) | | len(X) | → | n__len(X) |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__s(X)) | → | s(X) |
| add(0, X) | → | X | | add(X1, X2) | → | n__add(X1, X2) |
| len(nil) | → | 0 | | fst(0, Z) | → | nil |
| fst(X1, X2) | → | n__fst(X1, X2) | | s(X) | → | n__s(X) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) | | add(s(X), Y) | → | s(n__add(activate(X), Y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__add(X1, X2)) | → | activate#(X1) | | activate#(n__add(X1, X2)) | → | activate#(X2) |
| add#(s(X), Y) | → | activate#(X) |
Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__fst(X1, X2)) | → | activate#(X2) | | activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) |
| activate#(n__fst(X1, X2)) | → | activate#(X1) | | activate#(n__add(X1, X2)) | → | add#(activate(X1), activate(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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, nil, cons
Strategy
The following SCCs where found
| activate#(n__fst(X1, X2)) → activate#(X2) | activate#(n__fst(X1, X2)) → fst#(activate(X1), activate(X2)) |
| activate#(n__fst(X1, X2)) → activate#(X1) | fst#(s(X), cons(Y, Z)) → activate#(Z) |
| fst#(s(X), cons(Y, Z)) → activate#(X) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__fst(X1, X2)) | → | activate#(X2) | | activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) |
| activate#(n__fst(X1, X2)) | → | activate#(X1) | | 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(n__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) | | s(X) | → | n__s(X) |
| add(X1, X2) | → | n__add(X1, X2) | | len(X) | → | n__len(X) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(X) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__from, from, len, n__len, add, n__fst, fst, activate, n__s, 0, s, n__add, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): x
- add(x,y): y
- cons(x,y): y
- from(x): 0
- fst(x,y): 2y + 2x + 2
- fst#(x,y): 2y + x + 1
- len(x): 0
- n__add(x,y): y
- n__from(x): 0
- n__fst(x,y): 2y + 2x + 2
- n__len(x): 0
- n__s(x): x
- nil: 1
- s(x): x
Improved Usable rules
| from(X) | → | cons(X, n__from(n__s(X))) | | len(X) | → | n__len(X) |
| len(cons(X, Z)) | → | s(n__len(activate(Z))) | | activate(n__add(X1, X2)) | → | add(activate(X1), activate(X2)) |
| activate(n__fst(X1, X2)) | → | fst(activate(X1), activate(X2)) | | activate(n__s(X)) | → | s(X) |
| add(0, X) | → | X | | add(X1, X2) | → | n__add(X1, X2) |
| len(nil) | → | 0 | | fst(0, Z) | → | nil |
| fst(X1, X2) | → | n__fst(X1, X2) | | s(X) | → | n__s(X) |
| activate(n__len(X)) | → | len(activate(X)) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| fst(s(X), cons(Y, Z)) | → | cons(Y, n__fst(activate(X), activate(Z))) | | add(s(X), Y) | → | s(n__add(activate(X), Y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__fst(X1, X2)) | → | activate#(X2) | | activate#(n__fst(X1, X2)) | → | fst#(activate(X1), activate(X2)) |
| activate#(n__fst(X1, X2)) | → | activate#(X1) | | fst#(s(X), cons(Y, Z)) | → | activate#(X) |
| fst#(s(X), cons(Y, Z)) | → | activate#(Z) |