Termination w.r.t. Q proof of /tmp/tmpJgmJcO/Ex8_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:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FIB(N) → FIB1(s(0), s(0))
ACTIVATE(n__fib1(X1, X2)) → FIB1(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X2)
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X1)
ADD(s(X), Y) → ADD(X, Y)
FIB(N) → SEL(N, fib1(s(0), s(0)))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(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:

FIB(N) → FIB1(s(0), s(0))
ACTIVATE(n__fib1(X1, X2)) → FIB1(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X2)
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X1)
ADD(s(X), Y) → ADD(X, Y)
FIB(N) → SEL(N, fib1(s(0), s(0)))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

ADD(s(X), Y) → ADD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

From the DPs we obtained the following set of size-change graphs:

ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X2)
The graph contains the following edges 1 > 1
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
The graph contains the following edges 1 > 1
ACTIVATE(n__fib1(X1, X2)) → ACTIVATE(X1)
The graph contains the following edges 1 > 1
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPSizeChangeProof

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
The graph contains the following edges 1 > 1