YES
The TRS could be proven terminating. The proof took 16 ms.
Problem 1 was processed with processor DependencyGraph (3ms). | Problem 2 was processed with processor SubtermCriterion (1ms).
sum#(s(x)) | → | sqr#(s(x)) | sum#(s(x)) | → | sum#(x) |
sum(0) | → | 0 | sum(s(x)) | → | +(sqr(s(x)), sum(x)) | |
sqr(x) | → | *(x, x) | sum(s(x)) | → | +(*(s(x), s(x)), sum(x)) |
Termination of terms over the following signature is verified: 0, s, sqr, *, sum, +
sum#(s(x)) → sum#(x) |
sum#(s(x)) | → | sum#(x) |
sum(0) | → | 0 | sum(s(x)) | → | +(sqr(s(x)), sum(x)) | |
sqr(x) | → | *(x, x) | sum(s(x)) | → | +(*(s(x), s(x)), sum(x)) |
Termination of terms over the following signature is verified: 0, s, sqr, *, sum, +
The following projection was used:
Thus, the following dependency pairs are removed:
sum#(s(x)) | → | sum#(x) |