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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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:

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
LEN(cons(X, Z)) → S(n__len(activate(Z)))
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), Y))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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:

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
LEN(cons(X, Z)) → S(n__len(activate(Z)))
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), Y))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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:

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → S(n__len(activate(Z)))
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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 4 less nodes.

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

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

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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.


FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__len(X)) → LEN(activate(X))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
Used ordering: Combined order from the following AFS and order.
FST(x1, x2)  =  FST(x1, x2)
s(x1)  =  x1
cons(x1, x2)  =  x2
ACTIVATE(x1)  =  x1
n__add(x1, x2)  =  n__add(x1, x2)
n__fst(x1, x2)  =  n__fst(x1, x2)
ADD(x1, x2)  =  x1
activate(x1)  =  x1
n__len(x1)  =  x1
LEN(x1)  =  x1
n__from(x1)  =  x1
fst(x1, x2)  =  fst(x1, x2)
add(x1, x2)  =  add(x1, x2)
0  =  0
from(x1)  =  x1
nil  =  nil
len(x1)  =  x1
n__s(x1)  =  x1

Recursive Path Order [2].
Precedence:
[nadd2, add2]
[nfst2, fst2] > FST2
[nfst2, fst2] > nil > 0


The following usable rules [14] were oriented:

fst(X1, X2) → n__fst(X1, X2)
add(0, X) → X
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
from(X) → n__from(X)
add(s(X), Y) → s(n__add(activate(X), Y))
fst(0, Z) → nil
activate(n__len(X)) → len(activate(X))
len(X) → n__len(X)
activate(n__s(X)) → s(X)
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
len(nil) → 0
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
add(X1, X2) → n__add(X1, X2)
activate(n__from(X)) → from(activate(X))
from(X) → cons(X, n__from(n__s(X)))
activate(X) → X
len(cons(X, Z)) → s(n__len(activate(Z)))
s(X) → n__s(X)



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

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

ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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 1 less node.

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

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

ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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.


ACTIVATE(n__from(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__len(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__len(x1)  =  x1
n__from(x1)  =  n__from(x1)
LEN(x1)  =  x1
cons(x1, x2)  =  x2
activate(x1)  =  x1
fst(x1, x2)  =  fst
n__fst(x1, x2)  =  n__fst
add(x1, x2)  =  x2
from(x1)  =  from(x1)
s(x1)  =  s
nil  =  nil
len(x1)  =  x1
n__s(x1)  =  n__s
0  =  0
n__add(x1, x2)  =  x2

Recursive Path Order [2].
Precedence:
[nfrom1, from1] > [s, ns]
[fst, nfst] > nil > 0 > [s, ns]


The following usable rules [14] were oriented:

fst(X1, X2) → n__fst(X1, X2)
add(0, X) → X
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
from(X) → n__from(X)
add(s(X), Y) → s(n__add(activate(X), Y))
fst(0, Z) → nil
activate(n__len(X)) → len(activate(X))
len(X) → n__len(X)
activate(n__s(X)) → s(X)
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
len(nil) → 0
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
add(X1, X2) → n__add(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(X) → X
from(X) → cons(X, n__from(n__s(X)))
len(cons(X, Z)) → s(n__len(activate(Z)))
s(X) → n__s(X)



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

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

ACTIVATE(n__len(X)) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → LEN(activate(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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.


ACTIVATE(n__len(X)) → ACTIVATE(X)
ACTIVATE(n__len(X)) → LEN(activate(X))
The remaining pairs can at least be oriented weakly.

LEN(cons(X, Z)) → ACTIVATE(Z)
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__len(x1)  =  n__len(x1)
LEN(x1)  =  x1
cons(x1, x2)  =  x2
activate(x1)  =  x1
fst(x1, x2)  =  fst
n__fst(x1, x2)  =  n__fst
add(x1, x2)  =  x2
from(x1)  =  from
n__from(x1)  =  n__from
s(x1)  =  s
nil  =  nil
len(x1)  =  len(x1)
n__s(x1)  =  n__s
0  =  0
n__add(x1, x2)  =  x2

Recursive Path Order [2].
Precedence:
[nlen1, len1] > 0 > [s, ns]
[fst, nfst] > nil > [s, ns]
[from, nfrom] > [s, ns]


The following usable rules [14] were oriented:

fst(X1, X2) → n__fst(X1, X2)
add(0, X) → X
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
from(X) → n__from(X)
add(s(X), Y) → s(n__add(activate(X), Y))
fst(0, Z) → nil
activate(n__len(X)) → len(activate(X))
len(X) → n__len(X)
activate(n__s(X)) → s(X)
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
len(nil) → 0
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
add(X1, X2) → n__add(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(X) → X
from(X) → cons(X, n__from(n__s(X)))
len(cons(X, Z)) → s(n__len(activate(Z)))
s(X) → n__s(X)



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

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

LEN(cons(X, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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 0 SCCs with 1 less node.