YES
The TRS could be proven terminating. The proof took 21 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__h(X)) | → | h#(activate(X)) | | activate#(n__f(X)) | → | f#(activate(X)) |
| activate#(n__h(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |
Rewrite Rules
| f(X) | → | g(n__h(n__f(X))) | | h(X) | → | n__h(X) |
| f(X) | → | n__f(X) | | activate(n__h(X)) | → | h(activate(X)) |
| activate(n__f(X)) | → | f(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, n__h, n__f, h
Strategy
The following SCCs where found
| activate#(n__h(X)) → activate#(X) | activate#(n__f(X)) → activate#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__h(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |
Rewrite Rules
| f(X) | → | g(n__h(n__f(X))) | | h(X) | → | n__h(X) |
| f(X) | → | n__f(X) | | activate(n__h(X)) | → | h(activate(X)) |
| activate(n__f(X)) | → | f(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, n__h, n__f, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__f(X)) | → | activate#(X) | | activate#(n__h(X)) | → | activate#(X) |