YES
The TRS could be proven terminating. The proof took 30 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (14ms).
| – Problem 2 was processed with processor SubtermCriterion (3ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| after#(s(N), cons(X, XS)) | → | after#(N, activate(XS)) | | activate#(n__from(X)) | → | from#(activate(X)) |
| activate#(n__s(X)) | → | activate#(X) | | after#(s(N), cons(X, XS)) | → | activate#(XS) |
| activate#(n__from(X)) | → | activate#(X) | | activate#(n__s(X)) | → | s#(activate(X)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | after(0, XS) | → | XS |
| after(s(N), cons(X, XS)) | → | after(N, activate(XS)) | | 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: n__s, after, activate, 0, n__from, s, from, cons
Strategy
The following SCCs where found
| after#(s(N), cons(X, XS)) → after#(N, activate(XS)) |
| activate#(n__s(X)) → activate#(X) | activate#(n__from(X)) → activate#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| after#(s(N), cons(X, XS)) | → | after#(N, activate(XS)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | after(0, XS) | → | XS |
| after(s(N), cons(X, XS)) | → | after(N, activate(XS)) | | 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: n__s, after, activate, 0, n__from, s, from, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| after#(s(N), cons(X, XS)) | → | after#(N, activate(XS)) |
Problem 3: SubtermCriterion
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))) | | after(0, XS) | → | XS |
| after(s(N), cons(X, XS)) | → | after(N, activate(XS)) | | 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: n__s, after, activate, 0, 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) |