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, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)



QTRS
  ↳ DependencyPairsProof

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

f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
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(x, c(x), c(y)) → F(y, x, y)
F(s(x), y, z) → F(x, s(c(y)), c(z))
F(x, c(x), c(y)) → F(y, y, f(y, x, y))

The TRS R consists of the following rules:

f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)

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(x, c(x), c(y)) → F(y, x, y)
F(s(x), y, z) → F(x, s(c(y)), c(z))
F(x, c(x), c(y)) → F(y, y, f(y, x, y))

The TRS R consists of the following rules:

f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)

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

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

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

F(s(x), y, z) → F(x, s(c(y)), c(z))

The TRS R consists of the following rules:

f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QReductionProof
          ↳ QDP

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

F(s(x), y, z) → F(x, s(c(y)), c(z))

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

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
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.

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)



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

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

F(s(x), y, z) → F(x, s(c(y)), c(z))

R is empty.
Q is empty.
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:



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

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

F(x, c(x), c(y)) → F(y, x, y)

The TRS R consists of the following rules:

f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y

The set Q consists of the following terms:

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QReductionProof

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

F(x, c(x), c(y)) → F(y, x, y)

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

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
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.

f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)



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

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

F(x, c(x), c(y)) → F(y, x, y)

R is empty.
Q is empty.
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: