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:

f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

f(x) → f(n__h(g(x)))
h(x) → n__h(x)
n__h(activate(x)) → h(x)
activate(x) → x

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

f(x) → f(n__h(g(x)))
h(x) → n__h(x)
n__h(activate(x)) → h(x)
activate(x) → x

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x₁)) = 2 + x₁   
POL(f(x₁)) = x₁   
POL(g(x₁)) = x₁   
POL(h(x₁)) = 1 + x₁   
POL(n__h(x₁)) = x₁   

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

h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X

(6) Obligation:

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

f(X) → g(n__h(f(X)))

Q is empty.

(7) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(8) Obligation:

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

f(X) → g(n__h(f(X)))

The set Q consists of the following terms:

f(x0)

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

F(X) → F(X)

The TRS R consists of the following rules:

f(X) → g(n__h(f(X)))

The set Q consists of the following terms:

f(x0)

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

(11) 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.

(12) Obligation:

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

F(X) → F(X)

R is empty.
The set Q consists of the following terms:

f(x0)

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

(13) 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].

f(x0)

(14) Obligation:

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

F(X) → F(X)

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

(15) 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.

(16) Obligation:

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

F(X) → F(X)

R is empty.
The set Q consists of the following terms:

f(x0)

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

(17) 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].

f(x0)

(18) Obligation:

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

F(X) → F(X)

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

(19) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = F(X) evaluates to t =F(X)

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

• Semiunifier: [ ]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(X) to F(X).

(20) FALSE