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) → g(X)
g(a) → f(b)
f(X) → h(a)
ab

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
ab

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(a) → F(b)
H(X) → G(X)
F(X) → A
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
ab

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

G(a) → F(b)
H(X) → G(X)
F(X) → A
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
ab

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

G(a) → F(b)
H(X) → G(X)
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
ab

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