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(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

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(s(x)) → F(g(x, x))
F(s(x)) → G(x, x)

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

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(s(x)) → F(g(x, x))
F(s(x)) → G(x, x)

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

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(s(x)) → F(g(x, x))
F(s(x)) → G(x, x)

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

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

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

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

F(s(x)) → F(g(x, x))

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
01

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