Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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:

CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
P(mark(X)) → P(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(p(s(X))) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(f(0)) → F(s(0))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(p(X)) → MARK(X)
MARK(f(X)) → MARK(X)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
F(active(X)) → F(X)
F(mark(X)) → F(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
ACTIVE(f(0)) → S(0)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
P(active(X)) → P(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(p(X)) → P(mark(X))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → CONS(0, f(s(0)))
MARK(0) → ACTIVE(0)
MARK(f(X)) → F(mark(X))
ACTIVE(f(s(0))) → P(s(0))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(f(s(0))) → F(p(s(0)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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:

CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
P(mark(X)) → P(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(p(s(X))) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(f(0)) → F(s(0))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(p(X)) → MARK(X)
MARK(f(X)) → MARK(X)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
F(active(X)) → F(X)
F(mark(X)) → F(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
ACTIVE(f(0)) → S(0)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
P(active(X)) → P(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(p(X)) → P(mark(X))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → CONS(0, f(s(0)))
MARK(0) → ACTIVE(0)
MARK(f(X)) → F(mark(X))
ACTIVE(f(s(0))) → P(s(0))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(f(s(0))) → F(p(s(0)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(active(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(active(X)) → P(X)
P(mark(X)) → P(X)

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:

P(active(X)) → P(X)
The graph contains the following edges 1 > 1
P(mark(X)) → P(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

From the DPs we obtained the following set of size-change graphs:

S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
S(active(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

CONS(X1, active(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

From the DPs we obtained the following set of size-change graphs:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

From the DPs we obtained the following set of size-change graphs:

F(mark(X)) → F(X)
The graph contains the following edges 1 > 1
F(active(X)) → F(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
ACTIVE(p(s(X))) → MARK(X)
MARK(f(X)) → MARK(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(f(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(f(x₁)) = 2 + 2·x₁
POL(mark(x₁)) = x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(p(s(X))) → MARK(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(p(s(X))) → MARK(X)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(f(x₁)) = 1
POL(mark(x₁)) = 0
POL(p(x₁)) = 1
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
p(active(X)) → p(X)
p(mark(X)) → p(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
ACTIVE(p(s(X))) → MARK(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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(p(X)) → MARK(X)
ACTIVE(p(s(X))) → MARK(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(f(x₁)) = 0
POL(mark(x₁)) = x₁
POL(p(x₁)) = 1 + x₁
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

mark(0) → active(0)
active(f(0)) → mark(cons(0, f(s(0))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(p(s(X))) → mark(X)
mark(s(X)) → active(s(mark(X)))
mark(f(X)) → active(f(mark(X)))
active(f(s(0))) → mark(f(p(s(0))))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
p(active(X)) → p(X)
p(mark(X)) → p(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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(p(X)) → ACTIVE(p(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 0
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(f(x₁)) = 0
POL(mark(x₁)) = x₁
POL(p(x₁)) = 1
POL(s(x₁)) = x₁

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
f(x1) = x1
s(x1) = s(x1)
0 = 0
MARK(x1) = MARK(x1)
p(x1) = x1
cons(x1, x2) = x1
mark(x1) = x1
active(x1) = x1

Recursive path order with status [2].
Quasi-Precedence:

[ACTIVE₁, 0, MARK₁] > s₁

Status:

MARK₁: multiset
s₁: multiset
0: multiset
ACTIVE₁: multiset

The following usable rules [17] were oriented:

f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
p(active(X)) → p(X)
p(mark(X)) → p(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(0) → active(0)
active(f(0)) → mark(cons(0, f(s(0))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(p(s(X))) → mark(X)
mark(s(X)) → active(s(mark(X)))
mark(f(X)) → active(f(mark(X)))
active(f(s(0))) → mark(f(p(s(0))))
mark(p(X)) → active(p(mark(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(f(0)) → MARK(cons(0, f(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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.

ACTIVE(f(0)) → MARK(cons(0, f(s(0))))

The remaining pairs can at least be oriented weakly.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
f(x1) = f(x1)
s(x1) = s(x1)
0 = 0
MARK(x1) = x1
p(x1) = x1
cons(x1, x2) = x1
mark(x1) = x1
active(x1) = x1

Recursive path order with status [2].
Quasi-Precedence:

[f₁, s₁] > 0

Status:

f₁: [1]
s₁: [1]
0: multiset

The following usable rules [17] were oriented:

f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
p(active(X)) → p(X)
p(mark(X)) → p(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(0) → active(0)
active(f(0)) → mark(cons(0, f(s(0))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(p(s(X))) → mark(X)
mark(s(X)) → active(s(mark(X)))
mark(f(X)) → active(f(mark(X)))
active(f(s(0))) → mark(f(p(s(0))))
mark(p(X)) → active(p(mark(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ AND

                                        ↳ QDP

                                          ↳ Instantiation

                                        ↳ QDP

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MARK(f(X)) → ACTIVE(f(mark(X))) we obtained the following new rules:

MARK(f(p(s(0)))) → ACTIVE(f(mark(p(s(0)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ AND

                                        ↳ QDP

                                          ↳ Instantiation

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                        ↳ QDP

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

ACTIVE(f(s(0))) → MARK(f(p(s(0))))
MARK(f(p(s(0)))) → ACTIVE(f(mark(p(s(0)))))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(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.

ACTIVE(f(s(0))) → MARK(f(p(s(0))))

The remaining pairs can at least be oriented weakly.

MARK(f(p(s(0)))) → ACTIVE(f(mark(p(s(0)))))

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

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( f(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	0	0	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( p(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

M( s(x₁) ) =

/	0	\
\	1	/

/	0	1	\
\	1	0	/

x₁

M( 0 ) =

/	1	\
\	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(mark(X)) → f(X)
f(active(X)) → f(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
p(active(X)) → p(X)
p(mark(X)) → p(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(0) → active(0)
active(f(0)) → mark(cons(0, f(s(0))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(p(s(X))) → mark(X)
mark(s(X)) → active(s(mark(X)))
mark(f(X)) → active(f(mark(X)))
active(f(s(0))) → mark(f(p(s(0))))
mark(p(X)) → active(p(mark(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ AND

                                        ↳ QDP

                                          ↳ Instantiation

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                        ↳ QDP

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

MARK(f(p(s(0)))) → ACTIVE(f(mark(p(s(0)))))

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ AND

                                        ↳ QDP

                                        ↳ QDP

                                          ↳ UsableRulesProof

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
mark(f(X)) → active(f(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(p(X)) → active(p(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
p(mark(X)) → p(X)
p(active(X)) → p(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ AND

                                        ↳ QDP

                                        ↳ QDP

                                          ↳ UsableRulesProof

                                            ↳ QDP

                                              ↳ QDPSizeChangeProof

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

MARK(cons(X1, X2)) → MARK(X1)

From the DPs we obtained the following set of size-change graphs:

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