YES
The TRS could be proven terminating. The proof took 27 ms.
Problem 1 was processed with processor DependencyGraph (7ms). | Problem 2 was processed with processor SubtermCriterion (0ms).
f#(a) | → | f#(c(a)) | f#(a) | → | f#(d(a)) | |
f#(c(a)) | → | f#(d(b)) | f#(c(b)) | → | f#(d(a)) | |
e#(g(X)) | → | e#(X) |
f(a) | → | f(c(a)) | f(c(X)) | → | X | |
f(c(a)) | → | f(d(b)) | f(a) | → | f(d(a)) | |
f(d(X)) | → | X | f(c(b)) | → | f(d(a)) | |
e(g(X)) | → | e(X) |
Termination of terms over the following signature is verified: f, g, d, e, b, c, a
e#(g(X)) → e#(X) |
e#(g(X)) | → | e#(X) |
f(a) | → | f(c(a)) | f(c(X)) | → | X | |
f(c(a)) | → | f(d(b)) | f(a) | → | f(d(a)) | |
f(d(X)) | → | X | f(c(b)) | → | f(d(a)) | |
e(g(X)) | → | e(X) |
Termination of terms over the following signature is verified: f, g, d, e, b, c, a
The following projection was used:
Thus, the following dependency pairs are removed:
e#(g(X)) | → | e#(X) |