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(x, a)) → f(a, f(f(a, x), f(a, a)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(a, f(x0, a))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(a, f(x0, a))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(a, f(x0, a))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(a, f(x0, a))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

F(a, f(x, a)) → F(a, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(a, f(x0, a))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F(a, f(x, a)) → F(a, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x2)
a  =  a
f(x1, x2)  =  f(x1, x2)

Recursive Path Order [2].
Precedence:
[F1, a, f2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

f(a, f(x0, a))

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