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
                  ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


I(t(x1)) → E(r(x1))
R(i(t(e(r(x1))))) → E(w(r(i(t(e(x1))))))
The remaining pairs can at least be oriented weakly.

E(w(x1)) → R(i(x1))
R(e(x1)) → W(r(x1))
W(r(x1)) → I(t(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(E(x1)) = (1/4)x_1   
POL(i(x1)) = (1/2)x_1   
POL(w(x1)) = x_1   
POL(t(x1)) = 1/2   
POL(e(x1)) = 1/4   
POL(r(x1)) = 1/4   
POL(R(x1)) = (1/2)x_1   
POL(I(x1)) = (1/4)x_1   
POL(W(x1)) = (1/2)x_1   
The value of delta used in the strict ordering is 1/16.
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
                ↳ QDP
                  ↳ QDPOrderProof
QDP
                      ↳ DependencyGraphProof

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

E(w(x1)) → R(i(x1))
R(e(x1)) → W(r(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 0 SCCs with 3 less nodes.