Termination w.r.t. Q proof of /tmp/tmp8uwloT/Ex6_9_Luc02c

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:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(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:

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X1)
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(cons1(X1, X2)) → MARK(X1)
A__2ND(cons(X, X1)) → MARK(X)

The TRS R consists of the following rules:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X1)
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(cons1(X1, X2)) → MARK(X1)
A__2ND(cons(X, X1)) → MARK(X)

The TRS R consists of the following rules:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(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.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X1)
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(cons1(X1, X2)) → MARK(X1)
A__2ND(cons(X, X1)) → MARK(X)

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

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	1	1	\
\	0	0	/

x₁

/	0	0	\
\	1	0	/

x₂

M( from(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	0	/

x₁

M( cons1(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

/	1	0	\
\	0	0	/

x₂

M( a__2nd(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	1	/

x₁

M( 2nd(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	1	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( a__from(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	0	/

x₁

M( s(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

Tuple symbols:

M( A__2ND(x₁) ) =

[

]

x₁

M( MARK(x₁) ) =

[

]

x₁

M( A__FROM(x₁) ) =

[

]

x₁

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:

mark(2nd(X)) → a__2nd(mark(X))
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
mark(from(X)) → a__from(mark(X))
a__from(X) → cons(mark(X), from(s(X)))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__from(X) → from(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
MARK(from(X)) → A__FROM(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))

The TRS R consists of the following rules:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(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.

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(2nd(X)) → A__2ND(mark(X))

The remaining pairs can at least be oriented weakly.

A__2ND(cons(X, X1)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))

Used ordering: Polynomial interpretation [25]:

POL(2nd(x₁)) = 1 + x₁
POL(A__2ND(x₁)) = x₁
POL(A__FROM(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__2nd(x₁)) = 1 + x₁
POL(a__from(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(cons1(x₁, x₂)) = x₁ + x₂
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

mark(2nd(X)) → a__2nd(mark(X))
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
mark(from(X)) → a__from(mark(X))
a__from(X) → cons(mark(X), from(s(X)))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__from(X) → from(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__2ND(cons(X, X1)) → MARK(X1)
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))

The TRS R consists of the following rules:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

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

MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QDPSizeChangeProof

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

MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)

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:

MARK(cons1(X1, X2)) → MARK(X2)
The graph contains the following edges 1 > 1
MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
MARK(cons1(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1