Termination w.r.t. Q proof of TRS/nontermin/AG01#4.12a.trs

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:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y

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(0, 1, x) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y

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(0, 1, x) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y

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

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

F(0, 1, x) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y

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