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:

f(g(X), Y) → f(X, f(g(X), Y))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

f(g(X), Y) → f(X, f(g(X), Y))

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(g(X), Y) → f(X, f(g(X), Y))

The set Q consists of the following terms:

f(g(x0), x1)


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(g(X), Y) → F(X, f(g(X), Y))
F(g(X), Y) → F(g(X), Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, f(g(X), Y))

The set Q consists of the following terms:

f(g(x0), x1)

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

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

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

F(g(X), Y) → F(X, f(g(X), Y))
F(g(X), Y) → F(g(X), Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, f(g(X), Y))

The set Q consists of the following terms:

f(g(x0), x1)

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(g(X), Y) → F(X, f(g(X), Y))
The remaining pairs can at least be oriented weakly.

F(g(X), Y) → F(g(X), Y)
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
g(x1)  =  g(x1)
f(x1, x2)  =  f

Lexicographic Path Order [19].
Precedence:
F2 > f > g1


The following usable rules [14] were oriented:

f(g(X), Y) → f(X, f(g(X), Y))



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

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

F(g(X), Y) → F(g(X), Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, f(g(X), Y))

The set Q consists of the following terms:

f(g(x0), x1)

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