Termination w.r.t. Q proof of /tmp/tmpkkOlz0/Ex1_GM99

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:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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:

ACTIVE(c) → MARK(a)
MARK(f(X1, X2, X3)) → MARK(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
F(X1, mark(X2), X3) → F(X1, X2, X3)
MARK(b) → ACTIVE(b)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
MARK(a) → ACTIVE(a)
F(X1, active(X2), X3) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
MARK(c) → ACTIVE(c)
ACTIVE(c) → MARK(b)
MARK(f(X1, X2, X3)) → F(X1, X2, mark(X3))
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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:

ACTIVE(c) → MARK(a)
MARK(f(X1, X2, X3)) → MARK(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
F(X1, mark(X2), X3) → F(X1, X2, X3)
MARK(b) → ACTIVE(b)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
MARK(a) → ACTIVE(a)
F(X1, active(X2), X3) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
MARK(c) → ACTIVE(c)
ACTIVE(c) → MARK(b)
MARK(f(X1, X2, X3)) → F(X1, X2, mark(X3))
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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 with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

F(X1, active(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)

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:

F(X1, active(X2), X3) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
F(X1, X2, active(X3)) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
F(mark(X1), X2, X3) → F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
F(X1, mark(X2), X3) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
F(X1, X2, mark(X3)) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
F(active(X1), X2, X3) → F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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(f(X1, X2, X3)) → MARK(X3)

The remaining pairs can at least be oriented weakly.

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))

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

M( active(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( c ) =

/	0	\
\	1	/

M( a ) =

/	0	\
\	0	/

M( f(x₁, ..., x₃) ) =

/	0	\
\	1	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

/	0	1	\
\	0	1	/

x₃

M( mark(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( b ) =

/	0	\
\	0	/

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(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:

f(X1, X2, active(X3)) → f(X1, X2, X3)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
active(c) → mark(b)
active(c) → mark(a)
mark(a) → active(a)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
active(f(a, b, X)) → mark(f(X, X, X))
mark(c) → active(c)
mark(b) → active(b)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, mark(X3)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(X1, X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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