YES
The TRS could be proven terminating. The proof took 1008 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (81ms).
| – Problem 2 was processed with processor SubtermCriterion (18ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (350ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (427ms).
| | | – Problem 5 was processed with processor DependencyGraph (4ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | sel#(s(X), cons(Y, Z)) | → | activate#(Z) |
| activate#(n__from(X)) | → | from#(activate(X)) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) | | activate#(n__s(X)) | → | s#(activate(X)) |
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) | | activate#(n__first(X1, X2)) | → | activate#(X2) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, activate, 0, n__from, s, n__first, from, first, sel, cons, nil
Strategy
The following SCCs where found
| activate#(n__first(X1, X2)) → first#(activate(X1), activate(X2)) | activate#(n__first(X1, X2)) → activate#(X1) |
| first#(s(X), cons(Y, Z)) → activate#(Z) | activate#(n__s(X)) → activate#(X) |
| activate#(n__from(X)) → activate#(X) | activate#(n__first(X1, X2)) → activate#(X2) |
| sel#(s(X), cons(Y, Z)) → sel#(X, activate(Z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, activate, 0, n__from, s, n__first, from, first, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) | | activate#(n__first(X1, X2)) | → | activate#(X2) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, activate, 0, n__from, s, n__first, from, first, sel, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): 2x
- cons(x,y): y
- first(x,y): y + x
- first#(x,y): 2y
- from(x): 2x + 1
- n__first(x,y): y + x
- n__from(x): 2x + 1
- n__s(x): x
- nil: 0
- s(x): x
- sel(x,y): 0
Improved Usable rules
| first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | cons(X, n__from(n__s(X))) |
| s(X) | → | n__s(X) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | first(0, Z) | → | nil |
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 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__first(X1, X2)) | → | activate#(X2) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__s, 0, s, n__from, n__first, from, sel, first, nil, cons
Strategy
Polynomial Interpretation
- 0: 1
- activate(x): x
- activate#(x): 2x
- cons(x,y): 2y
- first(x,y): y + x + 1
- first#(x,y): 2y
- from(x): 0
- n__first(x,y): y + x + 1
- n__from(x): 0
- n__s(x): 2x + 1
- nil: 2
- s(x): 2x + 1
- sel(x,y): 0
Improved Usable rules
| first(X1, X2) | → | n__first(X1, X2) | | from(X) | → | cons(X, n__from(n__s(X))) |
| s(X) | → | n__s(X) | | activate(X) | → | X |
| from(X) | → | n__from(X) | | activate(n__from(X)) | → | from(activate(X)) |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | first(0, Z) | → | nil |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__first(X1, X2)) | → | activate#(X2) |
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | activate#(Z) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, activate, 0, n__from, s, n__first, from, first, sel, cons, nil
Strategy
There are no SCCs!