Termination w.r.t. Q proof of /tmp/tmpkBj3A5/jambox6.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:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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:

A(y, c(b(a(0, x), 0))) → B(0, y)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(x, y) → A(c(y), a(0, x))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

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

A(y, c(b(a(0, x), 0))) → B(0, y)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(x, y) → A(c(y), a(0, x))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(x, y) → A(c(y), a(0, x)) we obtained the following new rules:

B(0, y_0) → A(c(y_0), a(0, 0))
B(y_3, 0) → A(c(0), a(0, y_3))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

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

A(y, c(b(a(0, x), 0))) → B(0, y)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(c(y_0), a(0, 0))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(0, y_0) → A(c(y_0), a(0, 0)) at position [1] we obtained the following new rules:

B(0, y0) → A(c(y0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

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

A(y, c(b(a(0, x), 0))) → B(0, y)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y0) → A(c(y0), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Instantiation

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

A(y, c(b(a(0, x), 0))) → B(0, y)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(c(0), a(0, y_3))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(x, y) → A(0, x) we obtained the following new rules:

B(y_3, 0) → A(0, y_3)
B(0, y_0) → A(0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Instantiation

                    ↳ QDP

                      ↳ DependencyGraphProof

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

B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → B(0, y)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(0, 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Instantiation

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(c(0), a(0, y_3))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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.

B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
B(y_3, 0) → A(c(0), a(0, y_3))

The remaining pairs can at least be oriented weakly.

A(y, c(b(a(0, x), 0))) → B(0, y)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)

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

M( c(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( b(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	1	1	/

x₁

/	0	0	\
\	0	1	/

x₂

M( a(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	1	/

x₂

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( B(x₁, x₂) ) =

[

]

x₁

[

]

x₂

M( A(x₁, x₂) ) =

[

]

x₁

[

]

x₂

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:

a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)
b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Instantiation

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Instantiation

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

A(y, c(b(a(0, x), 0))) → B(0, y)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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.