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(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
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:

ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__a) → A

The TRS R consists of the following rules:

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
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:

ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
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.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
activate1 > [ng1, f1, g1] > [nf1, a, na]

Status:
ng1: [1]
nf1: [1]
f1: [1]
a: []
na: []
g1: [1]
activate1: [1]


The following usable rules [FROCOS05] were oriented:

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
activate(X) → X

(6) Obligation:

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

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
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