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:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:

F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(a, f(a, x))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:

F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(a, f(a, x))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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
          ↳ DependencyGraphProof

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

F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(a, f(a, x))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:

B1(A(x)) → B1(B(x))
B1(A(x)) → B1(x)
B1(A(x)) → B1(B(B(x)))

The TRS R consists of the following rules:

B(A(x)) → B(B(B(x)))

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


The following pairs can be oriented strictly and are deleted.


F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
B1(x1)  =  x1
A(x1)  =  A(x1)
B(x1)  =  B

Recursive path order with status [2].
Quasi-Precedence:
A1 > B

Status:
trivial


The following usable rules [14] were oriented:

f(b, f(a, x)) → f(b, f(b, f(b, x)))



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:

A1(B(x)) → A1(x)
A1(B(x)) → A1(A(A(x)))
A1(B(x)) → A1(A(x))

The TRS R consists of the following rules:

A(B(x)) → A(A(A(x)))

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


The following pairs can be oriented strictly and are deleted.


F(a, f(b, x)) → F(a, x)
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A1(x1)  =  x1
B(x1)  =  B(x1)
A(x1)  =  A

Recursive path order with status [2].
Quasi-Precedence:
B1 > A

Status:
trivial


The following usable rules [14] were oriented:

f(a, f(b, x)) → f(a, f(a, f(a, x)))



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.