YES
The TRS could be proven terminating. The proof took 46 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (27ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
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)) | → | take#(X1, X2) |
| activate#(n__from(X)) | → | from#(X) | | sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
| sel#(s(N), cons(X, XS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(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) |
| take(X1, X2) | → | n__take(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__take(X1, X2)) | → | take(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 2nd, 0, s, n__from, take, from, head, n__take, sel, cons, nil
Strategy
The following SCCs where found
| take#(s(N), cons(X, XS)) → activate#(XS) | activate#(n__take(X1, X2)) → take#(X1, X2) |
| sel#(s(N), cons(X, XS)) → sel#(N, activate(XS)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(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) |
| take(X1, X2) | → | n__take(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__take(X1, X2)) | → | take(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 2nd, 0, s, n__from, take, from, head, n__take, 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)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| take#(s(N), cons(X, XS)) | → | activate#(XS) | | activate#(n__take(X1, X2)) | → | take#(X1, X2) |
Rewrite Rules
| from(X) | → | cons(X, n__from(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) |
| take(X1, X2) | → | n__take(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__take(X1, X2)) | → | take(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 2nd, 0, s, n__from, take, from, head, n__take, sel, cons, nil
Strategy
Projection
The following projection was used:
- π (activate#): 1
- π (take#): 2
Thus, the following dependency pairs are removed:
| take#(s(N), cons(X, XS)) | → | activate#(XS) | | activate#(n__take(X1, X2)) | → | take#(X1, X2) |