Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(0) Obligation:

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

F(f(X)) → C(n__f(g(n__f(X))))
C(X) → D(activate(X))
C(X) → ACTIVATE(X)
H(X) → C(n__d(X))
ACTIVATE(n__f(X)) → F(X)
ACTIVATE(n__d(X)) → D(X)

The TRS R consists of the following 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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

C(X) → ACTIVATE(X)
ACTIVATE(n__f(X)) → F(X)
F(f(X)) → C(n__f(g(n__f(X))))

The TRS R consists of the following 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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


C(X) → ACTIVATE(X)
ACTIVATE(n__f(X)) → F(X)
F(f(X)) → C(n__f(g(n__f(X))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
C(x1)  =  C(x1)
ACTIVATE(x1)  =  ACTIVATE(x1)
n__f(x1)  =  n__f(x1)
F(x1)  =  F(x1)
f(x1)  =  f(x1)
g(x1)  =  g(x1)
c(x1)  =  c(x1)
d(x1)  =  d
activate(x1)  =  activate(x1)
h(x1)  =  h(x1)
n__d(x1)  =  n__d

Lexicographic path order with status [LPO].
Precedence:
activate1 > f1 > C1 > ACTIVATE1 > F1 > nf1
activate1 > f1 > g1 > nf1
activate1 > f1 > c1 > d > nd > nf1
h1 > c1 > d > nd > nf1

Status:
C1: [1]
ACTIVATE1: [1]
nf1: [1]
F1: [1]
f1: [1]
g1: [1]
c1: [1]
d: []
activate1: [1]
h1: [1]
nd: []

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE