Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/TRCSRSLASHEx15_Luc98

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:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(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:

AND₂(true, X) -> ACTIVATE₁(X)
ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), activate₁(Y)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

AND₂(true, X) -> ACTIVATE₁(X)
ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), activate₁(Y)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(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 strictly oriented and are deleted.

ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)

The remaining pairs can at least by weakly be oriented.

FROM₁(X) -> ACTIVATE₁(X)

Used ordering: Combined order from the following AFS and order.
ADD₂(x1, x2) = ADD₂(x1, x2)
0 = 0
ACTIVATE₁(x1) = x1
FIRST₂(x1, x2) = FIRST₂(x1, x2)
s₁(x1) = s₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)
n__add₂(x1, x2) = n__add₂(x1, x2)
FROM₁(x1) = x1
n__first₂(x1, x2) = n__first₂(x1, x2)
n__from₁(x1) = n__from₁(x1)

Lexicographic Path Order [19].
Precedence:

0 > ADD₂
s₁ > ADD₂
cons₂ > ADD₂
n_{_add₂} > ADD₂
n_{_first₂} > FIRST₂ > ADD₂
n_{_from₁} > ADD₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

FROM₁(X) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(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.