Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
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(X, n__g(X), Y) → F(activate(Y), activate(Y), activate(Y))
F(X, n__g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
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 2 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) 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(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(6) 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)  =  ACTIVATE(x1)
n__g(x1)  =  n__g(x1)
f(x1, x2, x3)  =  x3
activate(x1)  =  x1
g(x1)  =  g(x1)
b  =  b
c  =  c

Recursive Path Order [RPO].
Precedence:
[ACTIVATE1, ng1, g1]
[b, c]


The following usable rules [FROCOS05] were oriented:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

(7) Obligation:

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

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(X, n__g(X), Y) → F(activate(Y), activate(Y), activate(Y))

The TRS R consists of the following rules:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

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