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(g(x), h(x), y, x) → f(y, y, y, x)
f(x, y, z, 0) → 2
g(0) → 2
h(0) → 2

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(g(x), h(x), y, x) → f(y, y, y, x)
f(x, y, z, 0) → 2
g(0) → 2
h(0) → 2

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:

F(g(x), h(x), y, x) → F(y, y, y, x)

The TRS R consists of the following rules:

f(g(x), h(x), y, x) → f(y, y, y, x)
f(x, y, z, 0) → 2
g(0) → 2
h(0) → 2

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:

F(g(x), h(x), y, x) → F(y, y, y, x)

The TRS R consists of the following rules:

f(g(x), h(x), y, x) → f(y, y, y, x)
f(x, y, z, 0) → 2
g(0) → 2
h(0) → 2

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 1 less node.