Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/TRCSRSLASHEx8_BLR02

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:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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:

SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
ACTIVATE₁(n__fib1₂(X1, X2)) -> FIB1₂(activate₁(X1), activate₁(X2))
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
FIB₁(N) -> SEL₂(N, fib1₂(s₁(0), s₁(0)))
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
FIB₁(N) -> FIB1₂(s₁(0), s₁(0))

The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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:

SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
ACTIVATE₁(n__fib1₂(X1, X2)) -> FIB1₂(activate₁(X1), activate₁(X2))
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
FIB₁(N) -> SEL₂(N, fib1₂(s₁(0), s₁(0)))
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
FIB₁(N) -> FIB1₂(s₁(0), s₁(0))

The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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(s₁(X), Y) -> ADD₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD₂(x1, x2) = ADD₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > ADD₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)

The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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.

ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__fib1₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = x1
n__add₂(x1, x2) = n__add₂(x1, x2)
n__fib1₂(x1, x2) = n__fib1₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
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.

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL₂(x1, x2) = SEL₁(x1)
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x2
activate₁(x1) = activate₁(x1)
n__fib1₂(x1, x2) = n__fib1
fib1₂(x1, x2) = fib1
n__add₂(x1, x2) = n__add₂(x1, x2)
add₂(x1, x2) = add₂(x1, x2)
0 = 0

Lexicographic Path Order [19].
Precedence:

SEL₁ > activate₁ > fib1 > n_{_fib1}
SEL₁ > activate₁ > add₂ > s₁ > n_{_fib1}
SEL₁ > activate₁ > add₂ > n_{_add₂} > n_{_fib1}
0 > n_{_fib1}

The following usable rules [14] were oriented:

activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
add₂(X1, X2) -> n__add₂(X1, X2)
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

fib₁(N) -> sel₂(N, fib1₂(s₁(0), s₁(0)))
fib1₂(X, Y) -> cons₂(X, n__fib1₂(Y, n__add₂(X, Y)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
fib1₂(X1, X2) -> n__fib1₂(X1, X2)
add₂(X1, X2) -> n__add₂(X1, X2)
activate₁(n__fib1₂(X1, X2)) -> fib1₂(activate₁(X1), activate₁(X2))
activate₁(n__add₂(X1, X2)) -> add₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.