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(X, g(X)) → f(1, g(X))
g(1) → g(0)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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(X, g(X)) → F(1, g(X))

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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