Termination w.r.t. Q proof of /tmp/tmpq6vKsy/kabasci03.trs

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:

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]:

POL(0) = 1
POL(H(x₁, x₂, x₃)) = x₁ + x₂
POL(c(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = x₁
POL(t(x₁)) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( c(x₁, x₂) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

/	1	0	\
\	0	1	/

x₂

M( t(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	0	0	/

x₁

M( 0 ) =

/	0	\
\	1	/

Tuple symbols:

M( H(x₁, ..., x₃) ) =

[

]

x₁

[

]

x₂

[

]

x₃

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

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

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.
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:

H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
The graph contains the following edges 2 > 2