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__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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__F1(s1(0)) -> A__P1(s1(0))
MARK1(s1(X)) -> MARK1(X)
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(s1(0)) -> A__F1(a__p1(s1(0)))
MARK1(p1(X)) -> A__P1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(p1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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:

A__F1(s1(0)) -> A__P1(s1(0))
MARK1(s1(X)) -> MARK1(X)
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(s1(0)) -> A__F1(a__p1(s1(0)))
MARK1(p1(X)) -> A__P1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(p1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

A__F1(s1(0)) -> A__F1(a__p1(s1(0)))

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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 strictly oriented and are deleted.


A__F1(s1(0)) -> A__F1(a__p1(s1(0)))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
A__F1(x1)  =  A__F1(x1)
s1(x1)  =  s
0  =  0
a__p1(x1)  =  a__p
p1(x1)  =  p

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


The following usable rules [14] were oriented:

a__p1(s1(0)) -> 0
a__p1(X) -> p1(X)



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

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

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

MARK1(s1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(f1(X)) -> MARK1(X)
MARK1(p1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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 strictly oriented and are deleted.


MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least by weakly be oriented.

MARK1(s1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)
MARK1(p1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
f1(x1)  =  x1
p1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

MARK1(s1(X)) -> MARK1(X)
MARK1(p1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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 strictly oriented and are deleted.


MARK1(p1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.

MARK1(s1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  x1
p1(x1)  =  p1(x1)
f1(x1)  =  x1

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


The following usable rules [14] were oriented: none



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

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

MARK1(s1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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 strictly oriented and are deleted.


MARK1(f1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.

MARK1(s1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  x1
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

MARK1(s1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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 strictly oriented and are deleted.


MARK1(s1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  s1(x1)

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


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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__f1(0) -> cons2(0, f1(s1(0)))
a__f1(s1(0)) -> a__f1(a__p1(s1(0)))
a__p1(s1(0)) -> 0
mark1(f1(X)) -> a__f1(mark1(X))
mark1(p1(X)) -> a__p1(mark1(X))
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__p1(X) -> p1(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.