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

The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)



↳ QTRS
  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)


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(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → H(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)

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(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → H(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

H(g(x, y)) → H(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

H(g(x, y)) → H(x)

R is empty.
The set Q consists of the following terms:

g(0, 1)
h(g(x0, x1))
f(0, 1, g(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

h(g(x0, x1))
f(0, 1, g(x0, x1), x2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPSizeChangeProof

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

H(g(x, y)) → H(x)

R is empty.
The set Q consists of the following terms:

g(0, 1)

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: