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:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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:

ADD2(s1(X), Y) -> ADD2(X, Y)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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:

ADD2(s1(X), Y) -> ADD2(X, Y)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))

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

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

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

FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)

The TRS R consists of the following rules:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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.


FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST2(x1, x2) ) = 3x2 + 2


POL( cons2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = max{0, -3}



The following usable rules [14] were oriented: none



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

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

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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
          ↳ QDP
            ↳ QDPOrderProof

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:

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( ADD2(x1, x2) ) = 2x1 + 3x2 + 2



The following usable rules [14] were oriented: none



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

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

and2(true, X) -> X
and2(false, Y) -> false
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(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.