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 oriented strictly and are deleted.


A__F1(s1(0)) -> A__F1(a__p1(s1(0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(A__F1(x1)) = 2·x1   
POL(a__p1(x1)) = 2   
POL(p1(x1)) = 0   
POL(s1(x1)) = 1 + 2·x1   

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 oriented strictly and are deleted.


MARK1(s1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(f1(X)) -> MARK1(X)
MARK1(p1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(cons2(x1, x2)) = x1   
POL(f1(x1)) = x1   
POL(p1(x1)) = x1   
POL(s1(x1)) = 1 + 2·x1   

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(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

MARK1(p1(X)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(cons2(x1, x2)) = 1 + 2·x1   
POL(f1(x1)) = x1   
POL(p1(x1)) = x1   

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(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 oriented strictly and are deleted.


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

MARK1(p1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(f1(x1)) = 1 + 2·x1   
POL(p1(x1)) = x1   

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(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 oriented strictly and are deleted.


MARK1(p1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = 2·x1   
POL(p1(x1)) = 1 + 2·x1   

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.