YES
The TRS could be proven terminating. The proof took 87 ms.
Problem 1 was processed with processor DependencyGraph (3ms). | Problem 2 was processed with processor PolynomialLinearRange4 (55ms).
c# | → | f#(g(c)) | T(c) | → | c# | |
T(g(x_1)) | → | T(x_1) |
c | → | f(g(c)) | f(g(X)) | → | g(X) |
Termination of terms over the following signature is verified: f, g, c
Context-sensitive strategy:
μ(f) = μ(T) = μ(g) = μ(f#) = μ(c) = μ(c#) = ∅
T(g(x_1)) → T(x_1) |
T(g(x_1)) | → | T(x_1) |
c | → | f(g(c)) | f(g(X)) | → | g(X) |
Termination of terms over the following signature is verified: f, g, c
Context-sensitive strategy:
μ(f) = μ(g) = μ(T) = μ(f#) = μ(c) = μ(c#) = ∅
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
T(g(x_1)) | → | T(x_1) |