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
  ↳ 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))

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:

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))

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:

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs.

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

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.


LEN1(cons2(X, Z)) -> LEN1(Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LEN1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 1 + 2·x2   

The following usable rules [14] were oriented: none



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

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
          ↳ 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))

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.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD2(x1, x2)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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
          ↳ 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))

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

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

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))

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.


FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FST2(x1, x2)) = 2·x1 + x2   
POL(cons2(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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.