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:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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:

H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(c(s(x), c(s(0), y)), z, t(x)) → T(c(x, s(x)))
H(x, c(y, z), t(w)) → T(c(t(w), w))
H(c(x, y), c(s(z), z), t(w)) → T(c(x, c(y, t(w))))
H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
H(c(x, y), c(s(z), z), t(w)) → T(t(c(x, c(y, t(w)))))
T(t(x)) → T(c(t(x), x))
H(c(s(x), c(s(0), y)), z, t(x)) → T(t(c(x, s(x))))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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:

H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(c(s(x), c(s(0), y)), z, t(x)) → T(c(x, s(x)))
H(x, c(y, z), t(w)) → T(c(t(w), w))
H(c(x, y), c(s(z), z), t(w)) → T(c(x, c(y, t(w))))
H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
H(c(x, y), c(s(z), z), t(w)) → T(t(c(x, c(y, t(w)))))
T(t(x)) → T(c(t(x), x))
H(c(s(x), c(s(0), y)), z, t(x)) → T(t(c(x, s(x))))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

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

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

H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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.


H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
The remaining pairs can at least be oriented weakly.

H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1, x2)) = 2 + x_1 + x_2   
POL(t(x1)) = 3   
POL(s(x1)) = x_1   
POL(0) = 0   
POL(H(x1, x2, x3)) = (4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



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

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

H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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