YES
The TRS could be proven terminating. The proof took 25 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |
| U11#(tt) | → | U12#(tt) | | isNePal#(__(I, __(P, I))) | → | U11#(tt) |
Rewrite Rules
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | | __(X, nil) | → | X |
| __(nil, X) | → | X | | U11(tt) | → | U12(tt) |
| U12(tt) | → | tt | | isNePal(__(I, __(P, I))) | → | U11(tt) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, tt, isNePal, __, U11, U12, nil
Strategy
The following SCCs where found
| __#(__(X, Y), Z) → __#(X, __(Y, Z)) | __#(__(X, Y), Z) → __#(Y, Z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |
Rewrite Rules
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | | __(X, nil) | → | X |
| __(nil, X) | → | X | | U11(tt) | → | U12(tt) |
| U12(tt) | → | tt | | isNePal(__(I, __(P, I))) | → | U11(tt) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, tt, isNePal, __, U11, U12, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |