(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(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.

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__fst(x1, x2)  =  n__fst(x1, x2)
FST(x1, x2)  =  FST(x1, x2)
activate(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  x2
n__from(x1)  =  x1
n__add(x1, x2)  =  n__add(x1, x2)
ADD(x1, x2)  =  x1
n__len(x1)  =  x1
LEN(x1)  =  x1
fst(x1, x2)  =  fst(x1, x2)
0  =  0
nil  =  nil
add(x1, x2)  =  add(x1, x2)
from(x1)  =  x1
n__s(x1)  =  x1
len(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nfst2, fst2] > FST2
[nfst2, fst2] > nil > 0
[nadd2, add2]

Status:
add2: [1,2]
nadd2: [1,2]
nfst2: [2,1]
fst2: [2,1]


The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nfrom1, from1] > [s, ns]
[fst, nfst] > nil > 0 > [s, ns]

Status:
trivial


The following usable rules [FROCOS05] were oriented:

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

(10) Obligation:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__len(x1)  =  n__len(x1)
LEN(x1)  =  x1
activate(x1)  =  x1
cons(x1, x2)  =  x2
fst(x1, x2)  =  fst
nil  =  nil
add(x1, x2)  =  x2
s(x1)  =  s
n__fst(x1, x2)  =  n__fst
from(x1)  =  from
n__from(x1)  =  n__from
len(x1)  =  len(x1)
0  =  0
n__s(x1)  =  n__s
n__add(x1, x2)  =  x2

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nlen1, len1] > 0 > [s, ns]
[fst, nfst] > nil > [s, ns]
[from, nfrom] > [s, ns]

Status:
trivial


The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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.

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(14) TRUE