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

(0) Obligation:

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x₁)) = 1 + x₁   
POL(c) = 1   
POL(d) = 1   
POL(g(x₁)) = 1 + x₁   
POL(h(x₁)) = x₁   
POL(n__c) = 0   
POL(n__d) = 1   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

c → n__c
activate(n__d) → d
activate(X) → X

(2) Obligation:

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

Q is empty.

(3) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

d → n__d
activate(n__c) → c
c → d

The TRS R 2 is

g(X) → h(activate(X))
h(n__d) → g(n__c)

The signature Sigma is {g, h}

(4) Obligation:

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

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

G(X) → H(activate(X))
G(X) → ACTIVATE(X)
C → D
H(n__d) → G(n__c)
ACTIVATE(n__c) → C

The TRS R consists of the following rules:

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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

(9) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

g(x0)
c
h(n__d)
d
activate(n__c)

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

(11) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

g(x0)
h(n__d)

(12) Obligation:

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

The set Q consists of the following terms:

c
d
activate(n__c)

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

(13) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(14) Obligation:

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

H(n__d) → G(n__c)
G(X) → H(activate(X))

The TRS R consists of the following rules:

activate(n__c) → c
c → d
d → n__d

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

(15) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = G(n__c) evaluates to t =G(n__c)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

G(n__c) → H(activate(n__c))
with rule G(X) → H(activate(X)) at position [] and matcher [X / n__c]

H(activate(n__c)) → H(c)
with rule activate(n__c) → c at position [0] and matcher [ ]

H(c) → H(d)
with rule c → d at position [0] and matcher [ ]

H(d) → H(n__d)
with rule d → n__d at position [0] and matcher [ ]

H(n__d) → G(n__c)
with rule H(n__d) → G(n__c)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.

(16) NO