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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 PisEmptyProof (⇔)
↳12 TRUE

(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))
ADD(s(X), Y) → ACTIVATE(X)
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(x0, x1)  =  ACTIVATE(x0, x1)
FST(x0, x1, x2)  =  FST(x0, x2)
ADD(x0, x1, x2)  =  ADD(x1, x2)
LEN(x0, x1)  =  LEN(x0, x1)

Tags:
ACTIVATE has argument tags [8,1] and root tag 2
FST has argument tags [10,6,4] and root tag 0
ADD has argument tags [3,1,8] and root tag 3
LEN has argument tags [8,1] and root tag 2

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE
n__fst(x1, x2)  =  n__fst(x1, x2)
FST(x1, x2)  =  x1
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)  =  ADD
n__len(x1)  =  x1
LEN(x1)  =  LEN
fst(x1, x2)  =  fst(x1, x2)
from(x1)  =  x1
n__s(x1)  =  x1
add(x1, x2)  =  add(x1, x2)
len(x1)  =  x1
0  =  0
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
[nfst2, fst2] > [ACTIVATE, nadd2, LEN, add2]
ADD > [ACTIVATE, nadd2, LEN, add2]
[0, nil] > [ACTIVATE, nadd2, LEN, add2]

Status:
ACTIVATE: []
nfst2: multiset
nadd2: [1,2]
ADD: multiset
LEN: []
fst2: multiset
add2: [1,2]
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

ACTIVATE(n__from(X)) → 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) 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)
ACTIVATE(n__len(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(x0, x1)  =  ACTIVATE(x0, x1)
LEN(x0, x1)  =  LEN(x0, x1)

Tags:
ACTIVATE has argument tags [0,0] and root tag 0
LEN has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE
n__from(x1)  =  n__from(x1)
n__len(x1)  =  n__len(x1)
LEN(x1)  =  LEN
activate(x1)  =  activate(x1)
cons(x1, x2)  =  x2
n__fst(x1, x2)  =  n__fst
fst(x1, x2)  =  fst
from(x1)  =  from(x1)
n__s(x1)  =  n__s
s(x1)  =  s
n__add(x1, x2)  =  n__add(x1, x2)
add(x1, x2)  =  add(x1, x2)
len(x1)  =  len(x1)
0  =  0
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
[nfst, fst] > [ACTIVATE, nfrom1, nlen1, LEN, activate1, from1, len1] > add2 > [ns, s, nadd2]
[nfst, fst] > [ACTIVATE, nfrom1, nlen1, LEN, activate1, from1, len1] > 0 > nil > [ns, s, nadd2]

Status:
ACTIVATE: multiset
nfrom1: multiset
nlen1: multiset
LEN: multiset
activate1: multiset
nfst: []
fst: []
from1: multiset
ns: []
s: []
nadd2: multiset
add2: [1,2]
len1: multiset
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented: none

(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)

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__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(x0, x1)  =  ACTIVATE(x1)
LEN(x0, x1)  =  LEN(x0, x1)

Tags:
ACTIVATE has argument tags [0,3] and root tag 0
LEN has argument tags [1,0] and root tag 1

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE
n__len(x1)  =  n__len(x1)
LEN(x1)  =  LEN(x1)
activate(x1)  =  x1
cons(x1, x2)  =  x2
n__fst(x1, x2)  =  x2
fst(x1, x2)  =  x2
n__from(x1)  =  n__from(x1)
from(x1)  =  from(x1)
n__s(x1)  =  n__s
s(x1)  =  s
n__add(x1, x2)  =  n__add(x1, x2)
add(x1, x2)  =  add(x1, x2)
len(x1)  =  len(x1)
0  =  0
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
[ACTIVATE, nlen1, LEN1, len1] > 0 > [ns, s, nil]
[nfrom1, from1] > [ns, s, nil]
[nadd2, add2] > [ns, s, nil]

Status:
ACTIVATE: multiset
nlen1: [1]
LEN1: [1]
nfrom1: multiset
from1: multiset
ns: multiset
s: multiset
nadd2: [2,1]
add2: [2,1]
len1: [1]
0: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

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

(10) Obligation:

Q DP problem:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE