Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

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:

H(x, x, y) → G(x)
G(a) → H(a, b, a)
I(x) → F(x, x)

The TRS R consists of the following rules:

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

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:

H(x, x, y) → G(x)
G(a) → H(a, b, a)
I(x) → F(x, x)

The TRS R consists of the following rules:

f(x, y) → x
g(a) → h(a, b, a)
i(x) → f(x, x)
h(x, x, y) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.