NO

The TRS could be proven non-terminating. The proof took 337 ms.

The following reduction sequence is a witness for non-termination:

f#(___X) →* f#(___X)

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (9ms).
 | – Problem 2 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 3 was processed with processor BackwardInstantiation (1ms).
 |    |    | – Problem 4 was processed with processor BackwardInstantiation (1ms).
 |    |    |    | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (0ms)].

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

activate#(n__h(X))h#(X)f#(X)f#(X)

Rewrite Rules

f(X)g(n__h(f(X)))h(X)n__h(X)
activate(n__h(X))h(X)activate(X)X

Original Signature

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

Strategy


The following SCCs where found

f#(X) → f#(X)

Problem 2: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(X)f#(X)

Rewrite Rules

f(X)g(n__h(f(X)))h(X)n__h(X)
activate(n__h(X))h(X)activate(X)X

Original Signature

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

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(X) → f#(X) on dependency pair chains it holds that: Thus, f#(X) → f#(X) is replaced by instances determined through the above matching. These instances are:
f#(_X) → f#(_X)

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(_X)f#(_X)

Rewrite Rules

f(X)g(n__h(f(X)))h(X)n__h(X)
activate(n__h(X))h(X)activate(X)X

Original Signature

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

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(_X) → f#(_X) on dependency pair chains it holds that: Thus, f#(_X) → f#(_X) is replaced by instances determined through the above matching. These instances are:
f#(__X) → f#(__X)

Problem 4: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(__X)f#(__X)

Rewrite Rules

f(X)g(n__h(f(X)))h(X)n__h(X)
activate(n__h(X))h(X)activate(X)X

Original Signature

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

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(__X) → f#(__X) on dependency pair chains it holds that: Thus, f#(__X) → f#(__X) is replaced by instances determined through the above matching. These instances are:
f#(___X) → f#(___X)