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, X) → h(X, X)
h(0, X) → f(0, X, X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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



↳ QTRS
  ↳ DependencyPairsProof

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

f(0, 1, X) → h(X, X)
h(0, X) → f(0, X, X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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


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, X) → H(X, X)
H(0, X) → F(0, X, X)

The TRS R consists of the following rules:

f(0, 1, X) → h(X, X)
h(0, X) → f(0, X, X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof

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

F(0, 1, X) → H(X, X)
H(0, X) → F(0, X, X)

The TRS R consists of the following rules:

f(0, 1, X) → h(X, X)
h(0, X) → f(0, X, X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

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
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QReductionProof

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

F(0, 1, X) → H(X, X)
H(0, X) → F(0, X, X)

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QReductionProof
            ↳ QDP
              ↳ Instantiation

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

F(0, 1, X) → H(X, X)
H(0, X) → F(0, X, X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(0, 1, X) → H(X, X) we obtained the following new rules:

F(0, 1, 1) → H(1, 1)



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

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

F(0, 1, 1) → H(1, 1)
H(0, X) → F(0, X, X)

R is empty.
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 2 less nodes.