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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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)
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 remaining pairs can at least be oriented weakly.

ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(ACTIVATE1(x1)) = 1 + 2·x1   
POL(ADD2(x1, x2)) = 3 + 3·x1 + x1·x2 + 2·x2   
POL(FIRST2(x1, x2)) = 3 + 2·x1 + 3·x1·x2   
POL(FROM1(x1)) = 2 + 3·x1 + 2·x12   
POL(cons2(x1, x2)) = 3 + 3·x1 + x2   
POL(n__add2(x1, x2)) = 1 + 2·x1 + 2·x1·x2 + 2·x2   
POL(n__first2(x1, x2)) = 2 + x1 + 2·x1·x2 + 3·x2   
POL(n__from1(x1)) = 1 + 2·x1 + x12   
POL(s1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)

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 [13,14,18] contains 0 SCCs with 1 less node.