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:

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

The set Q consists of the following terms:

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



↳ QTRS
  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

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


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(x, y) → F(x, y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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(x, y) → F(x, y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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(x, y) → F(x, y, x)

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

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

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



↳ 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(x, y) → F(x, y, 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, 0) → H(0, 0)



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

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

H(x, y) → F(x, y, x)
F(0, 1, 0) → H(0, 0)

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

H(0, 0) → F(0, 0, 0)



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

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

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

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.