The TRS could be proven terminating. The proof took 20 ms.
Problem 1 was processed with processor SubtermCriterion (1ms). | – Problem 2 was processed with processor SubtermCriterion (0ms).
| ack#(s(x), s(y)) | → | ack#(x, ack(s(x), y)) | ack#(s(x), s(y)) | → | ack#(s(x), y) | |
| ack#(s(x), 0) | → | ack#(x, s(0)) |
| ack(0, y) | → | s(y) | ack(s(x), 0) | → | ack(x, s(0)) | |
| ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Termination of terms over the following signature is verified: 0, s, ack
The following projection was used:
Thus, the following dependency pairs are removed:
| ack#(s(x), s(y)) | → | ack#(x, ack(s(x), y)) | ack#(s(x), 0) | → | ack#(x, s(0)) |
| ack#(s(x), s(y)) | → | ack#(s(x), y) |
| ack(0, y) | → | s(y) | ack(s(x), 0) | → | ack(x, s(0)) | |
| ack(s(x), s(y)) | → | ack(x, ack(s(x), y)) |
Termination of terms over the following signature is verified: 0, s, ack
The following projection was used:
Thus, the following dependency pairs are removed:
| ack#(s(x), s(y)) | → | ack#(s(x), y) |