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__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.

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

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(s1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPAfsSolverProof

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

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(s1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(first2(X1, X2)) -> MARK1(X2)
Used argument filtering: MARK1(x1)  =  x1
first2(x1, x2)  =  first2(x1, x2)
A__FIRST2(x1, x2)  =  x2
mark1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  x1
A__FROM1(x1)  =  x1
cons2(x1, x2)  =  x1
a__first2(x1, x2)  =  a__first2(x1, x2)
a__from1(x1)  =  x1
0  =  0
nil  =  nil
Used ordering: Quasi Precedence: [first_2, a__first_2] > nil


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
QDP
          ↳ DependencyGraphProof

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

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPAfsSolverProof

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

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
Used argument filtering: MARK1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  from1(x1)
A__FROM1(x1)  =  x1
mark1(x1)  =  x1
cons2(x1, x2)  =  x1
first2(x1, x2)  =  x2
a__first2(x1, x2)  =  x2
a__from1(x1)  =  a__from1(x1)
0  =  0
nil  =  nil
Used ordering: Quasi Precedence: [from_1, a__from_1] > nil 0 > nil


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
QDP
                  ↳ DependencyGraphProof

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

MARK1(s1(X)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof
QDP
                      ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(cons2(X1, X2)) -> MARK1(X1)
Used argument filtering: MARK1(x1)  =  x1
s1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPAfsSolverProof
QDP
                          ↳ QDPAfsSolverProof

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__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MARK1(s1(X)) -> MARK1(X)
Used argument filtering: MARK1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPAfsSolverProof
                        ↳ QDP
                          ↳ QDPAfsSolverProof
QDP
                              ↳ PisEmptyProof

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

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(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.