Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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(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)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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

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

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

(5) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
n__len(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  cons(x2)
n__add(x1, x2)  =  n__add(x1)
n__fst(x1, x2)  =  n__fst(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• ACTIVATE(n__fst(X1, X2)) → FST(X1, X2) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 1 > 2

• ADD(s(X), Y) → ACTIVATE(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 >= 1

• LEN(cons(X, Z)) → ACTIVATE(Z) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• ACTIVATE(n__add(X1, X2)) → ADD(X1, X2) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1

• ACTIVATE(n__len(X)) → LEN(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 >= 1

• FST(s(X), cons(Y, Z)) → ACTIVATE(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 >= 1

• FST(s(X), cons(Y, Z)) → ACTIVATE(Z) (allowed arguments on rhs = {1})
  The graph contains the following edges 2 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(6) TRUE