Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

Q is empty.


↳ 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

Q is empty.

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

Q is empty.
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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(s(x), y, z) → F(x, s(c(y)), c(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = (3/4)x_1   
POL(s(x1)) = 1 + x_1   
POL(F(x1, x2, x3)) = (4)x_1 + (9/4)x_2 + (9/4)x_3   
The value of delta used in the strict ordering is 7/4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

F(x, c(x), c(y)) → F(y, x, y)
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

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