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

The set Q consists of the following terms:

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

The set Q consists of the following terms:

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) → G(X, X)
F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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, X) → G(X, X)
F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

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

F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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.

f(0, 1, x0)



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

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

F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

g(x0, x1)

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

F(0, 1, x0) → F(x0, x0, x0)



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

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

F(0, 1, x0) → F(x0, x0, x0)

The TRS R consists of the following rules:

g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

g(x0, x1)

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.