Termination w.r.t. Q proof of /tmp/tmpk3R65R/matchbox1.srs

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

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

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.

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

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

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

e(r(x)) → r(w(x))
t(i(x)) → r(e(x))
w(e(x)) → i(r(x))
e(t(x)) → e(r(x))
r(w(x)) → t(i(x))
r(e(x)) → w(e(x))
r(e(t(i(r(x))))) → e(t(i(r(w(e(x))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

e(r(x)) → r(w(x))
t(i(x)) → r(e(x))
w(e(x)) → i(r(x))
e(t(x)) → e(r(x))
r(w(x)) → t(i(x))
r(e(x)) → w(e(x))
r(e(t(i(r(x))))) → e(t(i(r(w(e(x))))))

Q is empty.

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

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

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

e(r(x)) → r(w(x))
t(i(x)) → r(e(x))
w(e(x)) → i(r(x))
e(t(x)) → e(r(x))
r(w(x)) → t(i(x))
r(e(x)) → w(e(x))
r(e(t(i(r(x))))) → e(t(i(r(w(e(x))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

e(r(x)) → r(w(x))
t(i(x)) → r(e(x))
w(e(x)) → i(r(x))
e(t(x)) → e(r(x))
r(w(x)) → t(i(x))
r(e(x)) → w(e(x))
r(e(t(i(r(x))))) → e(t(i(r(w(e(x))))))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

E(w(x1)) → R(i(x1))
E(r(x1)) → W(x1)
R(i(t(e(r(x1))))) → W(r(i(t(e(x1)))))
R(i(t(e(r(x1))))) → I(t(e(x1)))
R(e(x1)) → W(r(x1))
W(r(x1)) → T(x1)
E(w(x1)) → I(x1)
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
I(t(x1)) → R(x1)
R(i(t(e(r(x1))))) → T(e(x1))
W(r(x1)) → I(t(x1))
I(t(x1)) → E(r(x1))
R(e(x1)) → R(x1)
R(i(t(e(r(x1))))) → R(i(t(e(x1))))
R(i(t(e(r(x1))))) → E(x1)
E(r(x1)) → E(w(x1))
T(e(x1)) → R(e(x1))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

E(w(x1)) → R(i(x1))
E(r(x1)) → W(x1)
R(i(t(e(r(x1))))) → W(r(i(t(e(x1)))))
R(i(t(e(r(x1))))) → I(t(e(x1)))
R(e(x1)) → W(r(x1))
W(r(x1)) → T(x1)
E(w(x1)) → I(x1)
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
I(t(x1)) → R(x1)
R(i(t(e(r(x1))))) → T(e(x1))
W(r(x1)) → I(t(x1))
I(t(x1)) → E(r(x1))
R(e(x1)) → R(x1)
R(i(t(e(r(x1))))) → R(i(t(e(x1))))
R(i(t(e(r(x1))))) → E(x1)
E(r(x1)) → E(w(x1))
T(e(x1)) → R(e(x1))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

E(w(x1)) → I(x1)

The remaining pairs can at least be oriented weakly.

E(w(x1)) → R(i(x1))
E(r(x1)) → W(x1)
R(i(t(e(r(x1))))) → W(r(i(t(e(x1)))))
R(i(t(e(r(x1))))) → I(t(e(x1)))
R(e(x1)) → W(r(x1))
W(r(x1)) → T(x1)
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
I(t(x1)) → R(x1)
R(i(t(e(r(x1))))) → T(e(x1))
W(r(x1)) → I(t(x1))
I(t(x1)) → E(r(x1))
R(e(x1)) → R(x1)
R(i(t(e(r(x1))))) → R(i(t(e(x1))))
R(i(t(e(r(x1))))) → E(x1)
E(r(x1)) → E(w(x1))
T(e(x1)) → R(e(x1))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( E(x₁) ) = x₁ + 1

POL( i(x₁) ) = max{0, x₁ - 1}

POL( T(x₁) ) = x₁ + 1

POL( w(x₁) ) = x₁

POL( t(x₁) ) = x₁ + 1

POL( e(x₁) ) = x₁

POL( r(x₁) ) = x₁

POL( R(x₁) ) = x₁ + 1

POL( I(x₁) ) = x₁

POL( W(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

e(r(x1)) → e(w(x1))
e(w(x1)) → r(i(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))
i(t(x1)) → e(r(x1))
r(e(x1)) → w(r(x1))
w(r(x1)) → i(t(x1))
t(e(x1)) → r(e(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

E(w(x1)) → R(i(x1))
E(r(x1)) → W(x1)
R(i(t(e(r(x1))))) → W(r(i(t(e(x1)))))
R(i(t(e(r(x1))))) → I(t(e(x1)))
R(e(x1)) → W(r(x1))
W(r(x1)) → T(x1)
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
I(t(x1)) → R(x1)
R(i(t(e(r(x1))))) → T(e(x1))
W(r(x1)) → I(t(x1))
I(t(x1)) → E(r(x1))
R(e(x1)) → R(x1)
R(i(t(e(r(x1))))) → R(i(t(e(x1))))
R(i(t(e(r(x1))))) → E(x1)
E(r(x1)) → E(w(x1))
T(e(x1)) → R(e(x1))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

E(r(x1)) → W(x1)
R(i(t(e(r(x1))))) → W(r(i(t(e(x1)))))
R(i(t(e(r(x1))))) → I(t(e(x1)))
W(r(x1)) → T(x1)
I(t(x1)) → R(x1)
R(i(t(e(r(x1))))) → T(e(x1))
R(e(x1)) → R(x1)
R(i(t(e(r(x1))))) → R(i(t(e(x1))))
R(i(t(e(r(x1))))) → E(x1)
E(r(x1)) → E(w(x1))

The remaining pairs can at least be oriented weakly.

E(w(x1)) → R(i(x1))
R(e(x1)) → W(r(x1))
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
W(r(x1)) → I(t(x1))
I(t(x1)) → E(r(x1))
T(e(x1)) → R(e(x1))

Used ordering: Polynomial interpretation [25]:

POL(E(x₁)) = 1 + 2·x₁
POL(I(x₁)) = x₁
POL(R(x₁)) = 3 + 4·x₁
POL(T(x₁)) = 3 + 4·x₁
POL(W(x₁)) = 1 + 2·x₁
POL(e(x₁)) = 1 + 2·x₁
POL(i(x₁)) = x₁
POL(r(x₁)) = 3 + 4·x₁
POL(t(x₁)) = 7 + 8·x₁
POL(w(x₁)) = 1 + 2·x₁

The following usable rules [17] were oriented:

e(r(x1)) → e(w(x1))
e(w(x1)) → r(i(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))
i(t(x1)) → e(r(x1))
r(e(x1)) → w(r(x1))
w(r(x1)) → i(t(x1))
t(e(x1)) → r(e(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

I(t(x1)) → E(r(x1))
E(w(x1)) → R(i(x1))
R(e(x1)) → W(r(x1))
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
T(e(x1)) → R(e(x1))
W(r(x1)) → I(t(x1))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.
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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

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

E(w(x1)) → R(i(x1))
I(t(x1)) → E(r(x1))
R(e(x1)) → W(r(x1))
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
W(r(x1)) → I(t(x1))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule I(t(x1)) → E(r(x1)) at position [0] we obtained the following new rules:

I(t(i(t(e(r(x0)))))) → E(e(w(r(i(t(e(x0)))))))
I(t(e(x0))) → E(w(r(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

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

E(w(x1)) → R(i(x1))
R(e(x1)) → W(r(x1))
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
I(t(i(t(e(r(x0)))))) → E(e(w(r(i(t(e(x0)))))))
W(r(x1)) → I(t(x1))
I(t(e(x0))) → E(w(r(x0)))

The TRS R consists of the following rules:

r(e(x1)) → w(r(x1))
i(t(x1)) → e(r(x1))
e(w(x1)) → r(i(x1))
t(e(x1)) → r(e(x1))
w(r(x1)) → i(t(x1))
e(r(x1)) → e(w(x1))
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.i: 0
E: 0
w: 0
t: 1
e: 0
r: 0
R: 0
I: 0
W: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

I.1(t.0(i.1(t.0(e.0(r.1(x0)))))) → E.0(e.0(w.0(r.0(i.1(t.0(e.1(x0)))))))
E.0(w.0(x1)) → R.0(i.0(x1))
W.0(r.1(x1)) → I.1(t.1(x1))
R.0(e.1(x1)) → W.0(r.1(x1))
I.1(t.0(e.0(x0))) → E.0(w.0(r.0(x0)))
I.1(t.0(i.1(t.0(e.0(r.0(x0)))))) → E.0(e.0(w.0(r.0(i.1(t.0(e.0(x0)))))))
W.0(r.0(x1)) → I.1(t.0(x1))
W.0(r.1(x1)) → I.0(t.1(x1))
E.0(w.1(x1)) → R.0(i.1(x1))
W.0(r.0(x1)) → I.0(t.0(x1))
R.0(i.1(t.0(e.0(r.0(x1))))) → E.0(w.0(r.0(i.1(t.0(e.0(x1))))))
I.1(t.0(e.1(x0))) → E.0(w.0(r.1(x0)))
R.0(e.0(x1)) → W.0(r.0(x1))
R.0(i.1(t.0(e.0(r.1(x1))))) → E.0(w.0(r.0(i.1(t.0(e.1(x1))))))

The TRS R consists of the following rules:

r.0(e.0(x1)) → w.0(r.0(x1))
e.0(w.1(x1)) → r.0(i.1(x1))
t.0(e.0(x1)) → r.0(e.0(x1))
t.1(x0) → t.0(x0)
r.0(e.1(x1)) → w.0(r.1(x1))
e.0(r.1(x1)) → e.0(w.1(x1))
w.1(x0) → w.0(x0)
r.0(i.1(t.0(e.0(r.1(x1))))) → e.0(w.0(r.0(i.1(t.0(e.1(x1))))))
e.0(w.0(x1)) → r.0(i.0(x1))
w.0(r.0(x1)) → i.1(t.0(x1))
r.1(x0) → r.0(x0)
i.1(t.1(x1)) → e.0(r.1(x1))
e.0(r.0(x1)) → e.0(w.0(x1))
i.1(t.0(x1)) → e.0(r.0(x1))
t.0(e.1(x1)) → r.0(e.1(x1))
w.0(r.1(x1)) → i.1(t.1(x1))
e.1(x0) → e.0(x0)
r.0(i.1(t.0(e.0(r.0(x1))))) → e.0(w.0(r.0(i.1(t.0(e.0(x1))))))
i.1(x0) → i.0(x0)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

I.1(t.0(i.1(t.0(e.0(r.1(x0)))))) → E.0(e.0(w.0(r.0(i.1(t.0(e.1(x0)))))))
E.0(w.0(x1)) → R.0(i.0(x1))
W.0(r.1(x1)) → I.1(t.1(x1))
R.0(e.1(x1)) → W.0(r.1(x1))
I.1(t.0(e.0(x0))) → E.0(w.0(r.0(x0)))
I.1(t.0(i.1(t.0(e.0(r.0(x0)))))) → E.0(e.0(w.0(r.0(i.1(t.0(e.0(x0)))))))
W.0(r.0(x1)) → I.1(t.0(x1))
W.0(r.1(x1)) → I.0(t.1(x1))
E.0(w.1(x1)) → R.0(i.1(x1))
W.0(r.0(x1)) → I.0(t.0(x1))
R.0(i.1(t.0(e.0(r.0(x1))))) → E.0(w.0(r.0(i.1(t.0(e.0(x1))))))
I.1(t.0(e.1(x0))) → E.0(w.0(r.1(x0)))
R.0(e.0(x1)) → W.0(r.0(x1))
R.0(i.1(t.0(e.0(r.1(x1))))) → E.0(w.0(r.0(i.1(t.0(e.1(x1))))))

The TRS R consists of the following rules:

r.0(e.0(x1)) → w.0(r.0(x1))
e.0(w.1(x1)) → r.0(i.1(x1))
t.0(e.0(x1)) → r.0(e.0(x1))
t.1(x0) → t.0(x0)
r.0(e.1(x1)) → w.0(r.1(x1))
e.0(r.1(x1)) → e.0(w.1(x1))
w.1(x0) → w.0(x0)
r.0(i.1(t.0(e.0(r.1(x1))))) → e.0(w.0(r.0(i.1(t.0(e.1(x1))))))
e.0(w.0(x1)) → r.0(i.0(x1))
w.0(r.0(x1)) → i.1(t.0(x1))
r.1(x0) → r.0(x0)
i.1(t.1(x1)) → e.0(r.1(x1))
e.0(r.0(x1)) → e.0(w.0(x1))
i.1(t.0(x1)) → e.0(r.0(x1))
t.0(e.1(x1)) → r.0(e.1(x1))
w.0(r.1(x1)) → i.1(t.1(x1))
e.1(x0) → e.0(x0)
r.0(i.1(t.0(e.0(r.0(x1))))) → e.0(w.0(r.0(i.1(t.0(e.0(x1))))))
i.1(x0) → i.0(x0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 14 less nodes.