Termination w.r.t. Q proof of /tmp/tmpodydPC/ExConc_Zan97

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

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

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

f(X) → g(n__h(f(X)))
h(X) → n__h(X)

Q is empty.

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

f(X) → g(n__h(f(X)))
h(X) → n__h(X)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

h(X) → n__h(X)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

f(X) → g(n__h(f(X)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                  ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

f(x0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

                          ↳ NonTerminationProof

                  ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

F(X) → F(X)

The TRS R consists of the following rules:none

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: [ ]

Rewriting sequence

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

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

f(x0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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.