Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__h1(X)) -> H1(activate1(X))
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__h1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__h1(X)) -> H1(activate1(X))
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__h1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__h1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__h1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = x1   
POL(n__f1(x1)) = 1 + x1   
POL(n__h1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f1(X) -> g1(n__h1(n__f1(X)))
h1(X) -> n__h1(X)
f1(X) -> n__f1(X)
activate1(n__h1(X)) -> h1(activate1(X))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.