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:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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, n__cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
ACTIVATE1(n__from1(X)) -> FROM1(X)
PI1(X) -> FROM1(0)
SQUARE1(X) -> TIMES2(X, X)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSPOS2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
PI1(X) -> 2NDSPOS2(X, from1(0))
PLUS2(s1(X), Y) -> PLUS2(X, Y)
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
2NDSPOS2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
TIMES2(s1(X), Y) -> TIMES2(X, Y)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
FROM1(X) -> CONS2(X, n__from1(s1(X)))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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, n__cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
ACTIVATE1(n__from1(X)) -> FROM1(X)
PI1(X) -> FROM1(0)
SQUARE1(X) -> TIMES2(X, X)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSPOS2(s1(N), cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
PI1(X) -> 2NDSPOS2(X, from1(0))
PLUS2(s1(X), Y) -> PLUS2(X, Y)
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
2NDSPOS2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
TIMES2(s1(X), Y) -> TIMES2(X, Y)
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
FROM1(X) -> CONS2(X, n__from1(s1(X)))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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 3 SCCs with 11 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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) ) = 2x1 + 2


POL( PLUS2(x1, x2) ) = max{0, 2x1 + x2 - 2}



The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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

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, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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) ) = 2x1 + 2


POL( TIMES2(x1, x2) ) = max{0, 2x1 + x2 - 2}



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:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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

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

2NDSPOS2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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, n__cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
2NDSNEG2(s1(N), cons2(X, n__cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from1(x1) ) = 2


POL( n__from1(x1) ) = max{0, -1}


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


POL( 2NDSPOS2(x1, x2) ) = 2x1 + 2x2 + 2


POL( s1(x1) ) = x1 + 2


POL( cons2(x1, x2) ) = x2 + 2


POL( n__cons2(x1, x2) ) = x2 + 2


POL( activate1(x1) ) = x1 + 2



The following usable rules [14] were oriented:

cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
from1(X) -> cons2(X, n__from1(s1(X)))
activate1(X) -> X



↳ 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:

from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(posrecip1(activate1(Y)), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, n__cons2(Y, Z))) -> rcons2(negrecip1(activate1(Y)), 2ndspos2(N, activate1(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)
from1(X) -> n__from1(X)
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(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.