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__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A__F(s(0)) → A__P(s(0))
MARK(s(X)) → MARK(X)
MARK(f(X)) → A__F(mark(X))
A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → A__P(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

A__F(s(0)) → A__P(s(0))
MARK(s(X)) → MARK(X)
MARK(f(X)) → A__F(mark(X))
A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → A__P(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(p(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

A__F(s(0)) → A__P(s(0))
MARK(s(X)) → MARK(X)
MARK(f(X)) → A__F(mark(X))
A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → A__P(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(f(X)) → MARK(X)
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

A__F(s(0)) → A__F(a__p(s(0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A__F(s(0)) → A__F(a__p(s(0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A__F(x1)  =  A__F(x1)
s(x1)  =  s
0  =  0
a__p(x1)  =  a__p
p(x1)  =  p

Lexicographic Path Order [19].
Precedence:
AF1 > ap > 0 > p
s > ap > 0 > p

The following usable rules [14] were oriented:

a__p(s(0)) → 0
a__p(X) → p(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(f(X)) → MARK(X)
MARK(p(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
f(x1)  =  x1
p(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
cons2 > MARK1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(p(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(p(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  x1
p(x1)  =  x1
f(x1)  =  f(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


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(p(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  s(x1)
p(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
s1 > MARK1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof

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

MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(p(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
p(x1)  =  p(x1)

Lexicographic Path Order [19].
Precedence:
p1 > MARK1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.