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:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.

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

AND2(true, X) -> ACTIVATE1(X)
ADD2(0, X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
IF3(true, X, Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(n__add2(activate1(X), activate1(Y)))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Y)
ADD2(s1(X), Y) -> ACTIVATE1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
ADD2(s1(X), Y) -> ACTIVATE1(Y)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
FROM1(X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
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:

AND2(true, X) -> ACTIVATE1(X)
ADD2(0, X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
IF3(true, X, Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(n__add2(activate1(X), activate1(Y)))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Y)
ADD2(s1(X), Y) -> ACTIVATE1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
ADD2(s1(X), Y) -> ACTIVATE1(Y)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
FROM1(X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

ADD2(0, X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(Y)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
FROM1(X) -> ACTIVATE1(X)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Y)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.
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(0, X) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(Y)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Y)
Used argument filtering: ADD2(x1, x2)  =  ADD2(x1, x2)
0  =  0
ACTIVATE1(x1)  =  x1
FIRST2(x1, x2)  =  FIRST2(x1, x2)
s1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
n__add2(x1, x2)  =  n__add2(x1, x2)
FROM1(x1)  =  x1
n__first2(x1, x2)  =  n__first2(x1, x2)
n__from1(x1)  =  x1
Used ordering: Quasi Precedence: [ADD_2, n__add_2] n__first_2 > FIRST_2


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ DependencyGraphProof

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

ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
FROM1(X) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> FROM1(X)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPAfsSolverProof

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

FROM1(X) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> FROM1(X)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.
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.

ACTIVATE1(n__from1(X)) -> FROM1(X)
Used argument filtering: FROM1(x1)  =  x1
ACTIVATE1(x1)  =  x1
n__from1(x1)  =  n__from1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPAfsSolverProof
QDP
                      ↳ DependencyGraphProof

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

FROM1(X) -> ACTIVATE1(X)

The TRS R consists of the following rules:

and2(true, X) -> activate1(X)
and2(false, Y) -> false
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
add2(0, X) -> activate1(X)
add2(s1(X), Y) -> s1(n__add2(activate1(X), activate1(Y)))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(activate1(Y), n__first2(activate1(X), activate1(Z)))
from1(X) -> cons2(activate1(X), n__from1(n__s1(activate1(X))))
add2(X1, X2) -> n__add2(X1, X2)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.