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:

f4(g1(x), h1(x), y, x) -> f4(y, y, y, x)
f4(x, y, z, 0) -> 2
g1(0) -> 2
h1(0) -> 2

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f4(g1(x), h1(x), y, x) -> f4(y, y, y, x)
f4(x, y, z, 0) -> 2
g1(0) -> 2
h1(0) -> 2

Q is empty.

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

F4(g1(x), h1(x), y, x) -> F4(y, y, y, x)

The TRS R consists of the following rules:

f4(g1(x), h1(x), y, x) -> f4(y, y, y, x)
f4(x, y, z, 0) -> 2
g1(0) -> 2
h1(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:

F4(g1(x), h1(x), y, x) -> F4(y, y, y, x)

The TRS R consists of the following rules:

f4(g1(x), h1(x), y, x) -> f4(y, y, y, x)
f4(x, y, z, 0) -> 2
g1(0) -> 2
h1(0) -> 2

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.