YES
The TRS could be proven terminating. The proof took 116 ms.
Problem 1 was processed with processor DependencyGraph (4ms). | Problem 2 was processed with processor PolynomialLinearRange4 (58ms). | | Problem 3 was processed with processor PolynomialLinearRange4 (23ms).
T(f(X)) | → | f#(X) | T(h(x_1)) | → | T(x_1) | |
T(f(x_1)) | → | T(x_1) |
f(X) | → | g(h(f(X))) |
Termination of terms over the following signature is verified: f, g, h
Context-sensitive strategy:
μ(T) = μ(g) = ∅
μ(f) = μ(f#) = μ(h) = {1}
T(h(x_1)) → T(x_1) | T(f(x_1)) → T(x_1) |
T(h(x_1)) | → | T(x_1) | T(f(x_1)) | → | T(x_1) |
f(X) | → | g(h(f(X))) |
Termination of terms over the following signature is verified: f, g, h
Context-sensitive strategy:
μ(g) = μ(T) = ∅
μ(f) = μ(f#) = μ(h) = {1}
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(h(x_1)) | → | T(x_1) |
T(f(x_1)) | → | T(x_1) |
f(X) | → | g(h(f(X))) |
Termination of terms over the following signature is verified: f, g, h
Context-sensitive strategy:
μ(T) = μ(g) = ∅
μ(f) = μ(f#) = μ(h) = {1}
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(f(x_1)) | → | T(x_1) |