YES
The TRS could be proven terminating. The proof took 225 ms.
Problem 1 was processed with processor DependencyGraph (11ms). | Problem 2 was processed with processor PolynomialLinearRange4 (98ms). | | Problem 3 was processed with processor PolynomialLinearRange4 (18ms). | | | Problem 4 was processed with processor PolynomialLinearRange4 (17ms).
f#(f(X)) | → | c#(f(g(f(X)))) | T(d(x_1)) | → | T(x_1) | |
T(f(X)) | → | f#(X) | T(f(x_1)) | → | T(x_1) | |
T(f(g(f(X)))) | → | f#(g(f(X))) | h#(X) | → | c#(d(X)) | |
T(g(x_1)) | → | T(x_1) |
f(f(X)) | → | c(f(g(f(X)))) | c(X) | → | d(X) | |
h(X) | → | c(d(X)) |
Termination of terms over the following signature is verified: f, g, d, c, h
Context-sensitive strategy:
μ(T) = μ(g) = μ(d) = μ(c) = μ(c#) = ∅
μ(f) = μ(f#) = μ(h#) = μ(h) = {1}
T(d(x_1)) → T(x_1) | T(f(x_1)) → T(x_1) |
T(g(x_1)) → T(x_1) |
T(d(x_1)) | → | T(x_1) | T(f(x_1)) | → | T(x_1) | |
T(g(x_1)) | → | T(x_1) |
f(f(X)) | → | c(f(g(f(X)))) | c(X) | → | d(X) | |
h(X) | → | c(d(X)) |
Termination of terms over the following signature is verified: f, g, d, c, h
Context-sensitive strategy:
μ(g) = μ(T) = μ(d) = μ(c) = μ(c#) = ∅
μ(f) = μ(f#) = μ(h#) = 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(g(x_1)) | → | T(x_1) |
T(d(x_1)) | → | T(x_1) | T(f(x_1)) | → | T(x_1) |
f(f(X)) | → | c(f(g(f(X)))) | c(X) | → | d(X) | |
h(X) | → | c(d(X)) |
Termination of terms over the following signature is verified: f, g, d, c, h
Context-sensitive strategy:
μ(T) = μ(g) = μ(d) = μ(c) = μ(c#) = ∅
μ(f) = μ(f#) = μ(h#) = 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(d(x_1)) | → | T(x_1) |
T(f(x_1)) | → | T(x_1) |
f(f(X)) | → | c(f(g(f(X)))) | c(X) | → | d(X) | |
h(X) | → | c(d(X)) |
Termination of terms over the following signature is verified: f, g, d, c, h
Context-sensitive strategy:
μ(g) = μ(T) = μ(d) = μ(c) = μ(c#) = ∅
μ(f) = μ(f#) = μ(h#) = 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) |