Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(f, a(g, a(f, x))) → A(g, a(g, a(f, x)))
A(g, a(f, a(g, x))) → A(f, a(f, a(g, x)))
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(f, a(g, a(f, x))) → A(g, a(g, a(f, x)))
A(g, a(f, a(g, x))) → A(f, a(f, a(g, x)))
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(f, a(g, a(f, x))) → A(g, a(g, a(f, x)))
A(g, a(f, a(g, x))) → A(f, a(f, a(g, x)))
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

F1(G(F(x))) → G1(G(F(x)))
F1(G(F(x))) → F1(G(G(F(x))))
G1(F(G(x))) → G1(F(F(G(x))))
G1(F(G(x))) → F1(F(G(x)))

The TRS R consists of the following rules:

F(G(F(x))) → F(G(G(F(x))))
G(F(G(x))) → G(F(F(G(x))))

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


The following pairs can be oriented strictly and are deleted.


A(f, a(g, a(f, x))) → A(g, a(g, a(f, x)))
A(g, a(f, a(g, x))) → A(f, a(f, a(g, x)))
The remaining pairs can at least be oriented weakly.

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
G(x1)  =  G
F(x1)  =  F
G1(x1)  =  G1

Lexicographic Path Order [19].
Precedence:
G > G1 > F11
G > G1 > F

The following usable rules [14] were oriented:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))
A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

A(f, a(g, a(f, x))) → A(f, a(g, a(g, a(f, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

A(g, a(f, a(g, x))) → A(g, a(f, a(f, a(g, x))))

The TRS R consists of the following rules:

a(f, a(g, a(f, x))) → a(f, a(g, a(g, a(f, x))))
a(g, a(f, a(g, x))) → a(g, a(f, a(f, a(g, x))))

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