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

The TRS R consists of the following rules:

f(n__f(n__a)) → f(n__g(n__f(n__a)))
f(X) → n__f(X)
a → n__a
g(X) → n__g(X)
activate(n__f(X)) → f(X)
activate(n__a) → a
activate(n__g(X)) → g(activate(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 4 less nodes.

(4) Obligation:

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

ACTIVATE(n__g(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

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


ACTIVATE(n__g(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)
f(x1)  =  f
n__f(x1)  =  n__f
n__a  =  n__a
a  =  a
g(x1)  =  g(x1)
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Precedence:
activate1 > f > nf
activate1 > f > na
activate1 > a > na
activate1 > g1 > ng1

Status:
ng1: multiset
f: multiset
nf: multiset
na: multiset
a: multiset
g1: multiset
activate1: [1]

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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