YES
The TRS could be proven terminating. The proof took 52 ms.
Problem 1 was processed with processor SubtermCriterion (0ms).
fib#(s(s(x))) | → | fib#(x) | fib#(s(s(x))) | → | fib#(s(x)) |
fib(0) | → | 0 | fib(s(0)) | → | s(0) | |
fib(s(s(x))) | → | +(fib(s(x)), fib(x)) |
Termination of terms over the following signature is verified: 0, s, +, fib
The following projection was used:
Thus, the following dependency pairs are removed:
fib#(s(s(x))) | → | fib#(x) | fib#(s(s(x))) | → | fib#(s(x)) |