Termination w.r.t. Q proof of TRS/TRCSREx2_Luc03b

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

Lexicographic path order with status [19].
Quasi-Precedence:

[n_{_add₂}, add_2]
[n_{_fst₂}, fst_2] > FST₂
[n_{_fst₂}, fst_2] > nil > 0

Status:

n_{_add₂}: [1,2]
fst₂: [2,1]
add₂: [1,2]
n_{_fst₂}: [2,1]

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

Lexicographic path order with status [19].
Quasi-Precedence:

[n_{_from₁}, from_1] > [s, n_{_s]}
[fst, n_{_fst]} > nil > 0 > [s, n_{_s]}

Status:

trivial

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

Lexicographic path order with status [19].
Quasi-Precedence:

[n_{_len₁}, len_1] > 0 > [s, n_{_s]}
[fst, n_{_fst]} > nil > [s, n_{_s]}
[from, n_{_from]} > [s, n_{_s]}

Status:

trivial

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.