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(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

The signature Sigma is {c, a, f, g, b}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

The set Q consists of the following terms:

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


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(0, 1, x) → F(x, x, x)
F(x, y, z) → G(x, y, z)

The TRS R consists of the following rules:

f(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

The set Q consists of the following terms:

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

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

G(0, 1, x) → F(x, x, x)
F(x, y, z) → G(x, y, z)

The TRS R consists of the following rules:

f(x, y, z) → g(x, y, z)
g(0, 1, x) → f(x, x, x)
a → b
a → c

The set Q consists of the following terms:

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

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

G(0, 1, x) → F(x, x, x)
F(x, y, z) → G(x, y, z)

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

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

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(x0, x1, x2)
g(0, 1, x0)
a



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Instantiation

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

G(0, 1, x) → F(x, x, x)
F(x, y, z) → G(x, y, z)

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

F(z0, z0, z0) → G(z0, z0, z0)



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

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

G(0, 1, x) → F(x, x, x)
F(z0, z0, z0) → G(z0, z0, z0)

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.