Termination w.r.t. Q proof of /tmp/tmpupm8bn/ex4.trs

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(h(x)) → f(i(x))
f(i(x)) → a
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

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

f(h(x)) → f(i(x))
f(i(x)) → a
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(x0))

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

f(h(x)) → f(i(x))
f(i(x)) → a
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(x0))

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

f(i(x)) → a

Used ordering:
Polynomial interpretation [25]:

POL(a) = 2
POL(f(x₁)) = 1 + 2·x₁
POL(h(x₁)) = 1 + x₁
POL(i(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(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(h(x)) → F(i(x))
F(h(x)) → I(x)

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(h(x)) → F(i(x))
F(h(x)) → I(x)

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

              ↳ UsableRulesProof

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(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(h(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

                      ↳ MNOCProof

              ↳ UsableRulesProof

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

                      ↳ MNOCProof

                        ↳ QDP

                          ↳ NonTerminationProof

              ↳ UsableRulesProof

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

s = F(i(x')) evaluates to t =F(i(x'))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

F(i(x')) → F(h(x'))
with rule i(x'') → h(x'') at position [0] and matcher [x'' / x']

F(h(x')) → F(i(x'))
with rule F(h(x)) → F(i(x))

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.

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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)
f(h(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(h(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)

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