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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 UsableRulesReductionPairsProof (⇔)
↳6 QDP
↳7 Narrowing (⇔)
↳8 QDP
↳9 Instantiation (⇔)
↳10 QDP
↳11 UsableRulesReductionPairsProof (⇔)
↳12 QDP
↳13 ForwardInstantiation (⇔)
↳14 QDP
↳15 NonTerminationProof (⇔)
↳16 NO

(0) Obligation:

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

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
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:

H(X) → G(X, X)
G(a, X) → F(b, activate(X))
G(a, X) → ACTIVATE(X)
F(X, X) → H(a)
F(X, X) → A

The TRS R consists of the following rules:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
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 2 less nodes.

(4) Obligation:

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

G(a, X) → F(b, activate(X))
F(X, X) → H(a)
H(X) → G(X, X)

The TRS R consists of the following rules:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X

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

(5) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x1 + x2   
POL(G(x1, x2)) = x1 + x2   
POL(H(x1)) = 2·x1   
POL(a) = 0   
POL(activate(x1)) = x1   
POL(b) = 0   

(6) Obligation:

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

G(a, X) → F(b, activate(X))
F(X, X) → H(a)
H(X) → G(X, X)

The TRS R consists of the following rules:

a → b
activate(X) → X

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

(7) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(a, X) → F(b, activate(X)) at position [] we obtained the following new rules [LPAR04]:

G(a, x0) → F(b, x0)

(8) Obligation:

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

F(X, X) → H(a)
H(X) → G(X, X)
G(a, x0) → F(b, x0)

The TRS R consists of the following rules:

a → b
activate(X) → X

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

(9) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(X, X) → H(a) we obtained the following new rules [LPAR04]:

F(b, b) → H(a)

(10) Obligation:

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

H(X) → G(X, X)
G(a, x0) → F(b, x0)
F(b, b) → H(a)

The TRS R consists of the following rules:

a → b
activate(X) → X

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

(11) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

activate(X) → X
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1, x2)) = 1 + 2·x1 + x2   
POL(G(x1, x2)) = 1 + x1 + x2   
POL(H(x1)) = 1 + 2·x1   
POL(a) = 0   
POL(b) = 0   

(12) Obligation:

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

H(X) → G(X, X)
G(a, x0) → F(b, x0)
F(b, b) → H(a)

The TRS R consists of the following rules:

a → b

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

(13) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule H(X) → G(X, X) we obtained the following new rules [LPAR04]:

H(a) → G(a, a)

(14) Obligation:

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

G(a, x0) → F(b, x0)
F(b, b) → H(a)
H(a) → G(a, a)

The TRS R consists of the following rules:

a → b

Q is empty.
We have to consider all minimal (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 = F(b, a) evaluates to t =F(b, a)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

F(b, a) → F(b, b)
with rule a → b at position [1] and matcher [ ]

F(b, b) → H(a)
with rule F(b, b) → H(a) at position [] and matcher [ ]

H(a) → G(a, a)
with rule H(a) → G(a, a) at position [] and matcher [ ]

G(a, a) → F(b, a)
with rule G(a, x0) → F(b, x0)

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