YES
The TRS could be proven terminating. The proof took 748 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (76ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (403ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (55ms).
| | | – Problem 5 was processed with processor PolynomialLinearRange4iUR (22ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| take#(s(N), cons(X, XS)) | → | activate#(XS) | | 2nd#(cons(X, XS)) | → | head#(activate(XS)) |
| 2nd#(cons(X, XS)) | → | activate#(XS) | | activate#(n__take(X1, X2)) | → | activate#(X1) |
| activate#(n__from(X)) | → | from#(activate(X)) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) | | activate#(n__take(X1, X2)) | → | activate#(X2) |
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) | | activate#(n__s(X)) | → | s#(activate(X)) |
| activate#(n__take(X1, X2)) | → | take#(activate(X1), activate(X2)) | | sel#(s(N), cons(X, XS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(activate(XS)) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | take(X1, X2) | → | n__take(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, n__from, from, n__take, n__s, activate, 0, s, take, head, sel, cons, nil
Strategy
The following SCCs where found
| take#(s(N), cons(X, XS)) → activate#(XS) | activate#(n__take(X1, X2)) → activate#(X1) |
| activate#(n__s(X)) → activate#(X) | activate#(n__from(X)) → activate#(X) |
| activate#(n__take(X1, X2)) → activate#(X2) | activate#(n__take(X1, X2)) → take#(activate(X1), activate(X2)) |
| sel#(s(N), cons(X, XS)) → sel#(N, activate(XS)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| take#(s(N), cons(X, XS)) | → | activate#(XS) | | activate#(n__take(X1, X2)) | → | activate#(X1) |
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__from(X)) | → | activate#(X) |
| activate#(n__take(X1, X2)) | → | activate#(X2) | | activate#(n__take(X1, X2)) | → | take#(activate(X1), activate(X2)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(activate(XS)) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | take(X1, X2) | → | n__take(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, n__from, from, n__take, n__s, activate, 0, s, take, head, sel, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- 2nd(x): 0
- activate(x): x
- activate#(x): x
- cons(x,y): y
- from(x): x
- head(x): 0
- n__from(x): x
- n__s(x): x
- n__take(x,y): 2y + 2x + 2
- nil: 1
- s(x): x
- sel(x,y): 0
- take(x,y): 2y + 2x + 2
- take#(x,y): 2y + 2x + 1
Improved Usable rules
| take(0, XS) | → | nil | | take(X1, X2) | → | n__take(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)) | | take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(n__s(X)) | → | s(activate(X)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| take#(s(N), cons(X, XS)) | → | activate#(XS) | | activate#(n__take(X1, X2)) | → | activate#(X1) |
| activate#(n__take(X1, X2)) | → | activate#(X2) | | activate#(n__take(X1, X2)) | → | take#(activate(X1), activate(X2)) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__from(X)) | → | activate#(X) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(activate(XS)) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | take(X1, X2) | → | n__take(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, n__from, from, n__take, n__s, activate, 0, s, take, head, sel, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- 2nd(x): 0
- activate(x): 0
- activate#(x): 3x
- cons(x,y): 0
- from(x): 0
- head(x): 0
- n__from(x): x + 1
- n__s(x): 3x
- n__take(x,y): 0
- nil: 0
- s(x): 0
- sel(x,y): 0
- take(x,y): 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:
| activate#(n__from(X)) | → | activate#(X) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(activate(XS)) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | take(X1, X2) | → | n__take(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, n__from, from, n__take, n__s, activate, 0, s, take, head, sel, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- 2nd(x): 0
- activate(x): 0
- activate#(x): x + 1
- cons(x,y): 0
- from(x): 0
- head(x): 0
- n__from(x): 0
- n__s(x): x + 1
- n__take(x,y): 0
- nil: 0
- s(x): 0
- sel(x,y): 0
- take(x,y): 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:
| activate#(n__s(X)) | → | activate#(X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(activate(XS)) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, n__take(N, activate(XS))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | take(X1, X2) | → | n__take(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__take(X1, X2)) | → | take(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, n__from, from, n__take, n__s, activate, 0, s, take, head, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |