Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))


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

ADD2(s1(X), Y) -> ADD2(X, Y)
LEN1(cons2(X, Z)) -> LEN1(Z)
FROM1(X) -> FROM1(s1(X))
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

ADD2(s1(X), Y) -> ADD2(X, Y)
LEN1(cons2(X, Z)) -> LEN1(Z)
FROM1(X) -> FROM1(s1(X))
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

LEN1(cons2(X, Z)) -> LEN1(Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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.

LEN1(cons2(X, Z)) -> LEN1(Z)
Used argument filtering: LEN1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP

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

ADD2(s1(X), Y) -> ADD2(X, Y)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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.

ADD2(s1(X), Y) -> ADD2(X, Y)
Used argument filtering: ADD2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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.

FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
Used argument filtering: FST2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))

The set Q consists of the following terms:

fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))

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