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:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(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:

AND(true, X) → ACTIVATE(X)
ADD(0, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

AND(true, X) → ACTIVATE(X)
ADD(0, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

ADD(0, X) → ACTIVATE(X)
AND(true, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(Y)
IF(false, X, Y) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPOrderProof

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

ADD(0, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(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.


ADD(0, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  ADD(x1, x2)
0  =  0
ACTIVATE(x1)  =  ACTIVATE(x1)
FIRST(x1, x2)  =  FIRST(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
n__add(x1, x2)  =  n__add(x1, x2)
FROM(x1)  =  FROM(x1)
n__first(x1, x2)  =  n__first(x1, x2)
n__from(x1)  =  n__from(x1)

Lexicographic path order with status [19].
Precedence:
0 > ACTIVATE1
s1 > ACTIVATE1
cons2 > ACTIVATE1
nadd2 > ADD2 > ACTIVATE1
nfirst2 > FIRST2
nfrom1 > FROM1 > ACTIVATE1

Status:
nadd2: multiset
ACTIVATE1: multiset
nfrom1: multiset
FIRST2: multiset
0: multiset
s1: multiset
ADD2: multiset
FROM1: multiset
cons2: multiset
nfirst2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
QDP
                  ↳ PisEmptyProof

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(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.