YES

The TRS could be proven terminating. The proof took 157 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (12ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (123ms).
 |    | – Problem 3 was processed with processor DependencyGraph (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(f(X))c#(n__f(g(n__f(X))))activate#(n__d(X))d#(X)
activate#(n__f(X))f#(X)h#(X)c#(n__d(X))
c#(X)d#(activate(X))c#(X)activate#(X)

Rewrite Rules

f(f(X))c(n__f(g(n__f(X))))c(X)d(activate(X))
h(X)c(n__d(X))f(X)n__f(X)
d(X)n__d(X)activate(n__f(X))f(X)
activate(n__d(X))d(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, n__d, g, d, c, n__f, h

Strategy


The following SCCs where found

f#(f(X)) → c#(n__f(g(n__f(X))))activate#(n__f(X)) → f#(X)
c#(X) → activate#(X)

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(f(X))c#(n__f(g(n__f(X))))activate#(n__f(X))f#(X)
c#(X)activate#(X)

Rewrite Rules

f(f(X))c(n__f(g(n__f(X))))c(X)d(activate(X))
h(X)c(n__d(X))f(X)n__f(X)
d(X)n__d(X)activate(n__f(X))f(X)
activate(n__d(X))d(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, n__d, g, d, c, n__f, h

Strategy


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:

f#(f(X))c#(n__f(g(n__f(X))))

Problem 3: DependencyGraph



Dependency Pair Problem

Dependency Pairs

activate#(n__f(X))f#(X)c#(X)activate#(X)

Rewrite Rules

f(f(X))c(n__f(g(n__f(X))))c(X)d(activate(X))
h(X)c(n__d(X))f(X)n__f(X)
d(X)n__d(X)activate(n__f(X))f(X)
activate(n__d(X))d(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: n__d, activate, f, g, d, c, h, n__f

Strategy


There are no SCCs!