YES
The TRS could be proven terminating. The proof took 20 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (8ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__a) | → | a# | | activate#(n__g(X)) | → | activate#(X) |
| activate#(n__f(X)) | → | f#(X) | | activate#(n__g(X)) | → | g#(activate(X)) |
| f#(n__f(n__a)) | → | f#(n__g(n__f(n__a))) |
Rewrite Rules
| f(n__f(n__a)) | → | f(n__g(n__f(n__a))) | | f(X) | → | n__f(X) |
| a | → | n__a | | g(X) | → | n__g(X) |
| activate(n__f(X)) | → | f(X) | | activate(n__a) | → | a |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, n__a, a, n__f, n__g
Strategy
The following SCCs where found
| activate#(n__g(X)) → activate#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__g(X)) | → | activate#(X) |
Rewrite Rules
| f(n__f(n__a)) | → | f(n__g(n__f(n__a))) | | f(X) | → | n__f(X) |
| a | → | n__a | | g(X) | → | n__g(X) |
| activate(n__f(X)) | → | f(X) | | activate(n__a) | → | a |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, n__a, a, n__f, n__g
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__g(X)) | → | activate#(X) |