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:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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:

2NDSPOS2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSNEG2(N, Z)
PLUS2(s1(X), Y) -> PLUS2(X, Y)
PI1(X) -> 2NDSPOS2(X, from1(0))
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
2NDSNEG2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSPOS2(N, Z)
TIMES2(s1(X), Y) -> TIMES2(X, Y)
SQUARE1(X) -> TIMES2(X, X)
PI1(X) -> FROM1(0)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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:

2NDSPOS2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSNEG2(N, Z)
PLUS2(s1(X), Y) -> PLUS2(X, Y)
PI1(X) -> 2NDSPOS2(X, from1(0))
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
2NDSNEG2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSPOS2(N, Z)
TIMES2(s1(X), Y) -> TIMES2(X, Y)
SQUARE1(X) -> TIMES2(X, X)
PI1(X) -> FROM1(0)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)

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 with 4 less nodes.

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

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

PLUS2(s1(X), Y) -> PLUS2(X, Y)

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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.


PLUS2(s1(X), Y) -> PLUS2(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( PLUS2(x1, x2) ) = 2x1 + 3x2 + 2



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:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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
          ↳ QDP
          ↳ QDP

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

TIMES2(s1(X), Y) -> TIMES2(X, Y)

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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.


TIMES2(s1(X), Y) -> TIMES2(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( TIMES2(x1, x2) ) = 2x1 + 3x2 + 2



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:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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
          ↳ QDP

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

2NDSPOS2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSNEG2(N, Z)
2NDSNEG2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSPOS2(N, Z)

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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.


2NDSPOS2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSNEG2(N, Z)
2NDSNEG2(s1(N), cons2(X, cons2(Y, Z))) -> 2NDSPOS2(N, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 2NDSPOS2(x1, x2) ) = max{0, 3x1 - 3}


POL( s1(x1) ) = 3x1 + 3


POL( cons2(x1, x2) ) = 0


POL( 2NDSNEG2(x1, x2) ) = max{0, x1 - 1}



The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, 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
          ↳ 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:

from1(X) -> cons2(X, from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, Z))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, Z))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)

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