YES
The TRS could be proven terminating. The proof took 17 ms.
Problem 1 was processed with processor DependencyGraph (5ms). | Problem 2 was processed with processor SubtermCriterion (1ms).
| and#(tt, X) | → | activate#(X) | __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | |
| __#(__(X, Y), Z) | → | __#(Y, Z) |
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | __(X, nil) | → | X | |
| __(nil, X) | → | X | and(tt, X) | → | activate(X) | |
| isNePal(__(I, __(P, I))) | → | tt | activate(X) | → | X |
Termination of terms over the following signature is verified: activate, tt, isNePal, __, nil, and
| __#(__(X, Y), Z) → __#(X, __(Y, Z)) | __#(__(X, Y), Z) → __#(Y, Z) |
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | __#(__(X, Y), Z) | → | __#(Y, Z) |
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | __(X, nil) | → | X | |
| __(nil, X) | → | X | and(tt, X) | → | activate(X) | |
| isNePal(__(I, __(P, I))) | → | tt | activate(X) | → | X |
Termination of terms over the following signature is verified: activate, tt, isNePal, __, nil, and
The following projection was used:
Thus, the following dependency pairs are removed:
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | __#(__(X, Y), Z) | → | __#(Y, Z) |