YES
The TRS could be proven terminating. The proof took 60001 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (14ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__from(X)) | → | from#(activate(X)) | | activate#(n__s(X)) | → | activate#(X) |
| 2nd#(cons(X, X1)) | → | 2nd#(cons1(X, activate(X1))) | | activate#(n__from(X)) | → | activate#(X) |
| 2nd#(cons(X, X1)) | → | activate#(X1) | | activate#(n__s(X)) | → | s#(activate(X)) |
Rewrite Rules
| 2nd(cons1(X, cons(Y, Z))) | → | Y | | 2nd(cons(X, X1)) | → | 2nd(cons1(X, activate(X1))) |
| from(X) | → | cons(X, n__from(n__s(X))) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, activate, n__s, cons1, n__from, s, from, cons
Strategy
The following SCCs where found
| activate#(n__s(X)) → activate#(X) | activate#(n__from(X)) → activate#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__from(X)) | → | activate#(X) |
Rewrite Rules
| 2nd(cons1(X, cons(Y, Z))) | → | Y | | 2nd(cons(X, X1)) | → | 2nd(cons1(X, activate(X1))) |
| from(X) | → | cons(X, n__from(n__s(X))) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: 2nd, activate, n__s, cons1, n__from, s, from, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__from(X)) | → | activate#(X) |