YES

The TRS could be proven terminating. The proof took 116 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (4ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (58ms).
 |    | – Problem 3 was processed with processor PolynomialLinearRange4 (23ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

T(f(X))f#(X)T(h(x_1))T(x_1)
T(f(x_1))T(x_1)

Rewrite Rules

f(X)g(h(f(X)))

Original Signature

Termination of terms over the following signature is verified: f, g, h

Strategy

Context-sensitive strategy:
μ(T) = μ(g) = ∅
μ(f) = μ(f#) = μ(h) = {1}


The following SCCs where found

T(h(x_1)) → T(x_1)T(f(x_1)) → T(x_1)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(h(x_1))T(x_1)T(f(x_1))T(x_1)

Rewrite Rules

f(X)g(h(f(X)))

Original Signature

Termination of terms over the following signature is verified: f, g, h

Strategy

Context-sensitive strategy:
μ(g) = μ(T) = ∅
μ(f) = μ(f#) = μ(h) = {1}


Polynomial Interpretation

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)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(f(x_1))T(x_1)

Rewrite Rules

f(X)g(h(f(X)))

Original Signature

Termination of terms over the following signature is verified: f, g, h

Strategy

Context-sensitive strategy:
μ(T) = μ(g) = ∅
μ(f) = μ(f#) = μ(h) = {1}


Polynomial Interpretation

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)