The TRS could be proven terminating. The proof took 18 ms.
Problem 1 was processed with processor SubtermCriterion (1ms). | – Problem 2 was processed with processor DependencyGraph (0ms).
| f#(cons(x, k), l) | → | g#(k, l, cons(x, k)) | g#(a, b, c) | → | f#(a, cons(b, c)) |
| f(empty, l) | → | l | f(cons(x, k), l) | → | g(k, l, cons(x, k)) | |
| g(a, b, c) | → | f(a, cons(b, c)) |
Termination of terms over the following signature is verified: f, g, empty, cons
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(cons(x, k), l) | → | g#(k, l, cons(x, k)) |
| g#(a, b, c) | → | f#(a, cons(b, c)) |
| f(empty, l) | → | l | f(cons(x, k), l) | → | g(k, l, cons(x, k)) | |
| g(a, b, c) | → | f(a, cons(b, c)) |
Termination of terms over the following signature is verified: f, g, empty, cons