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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, 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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, X2)
activate1(X) -> X

Q is empty.

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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD2(s1(X), Y) -> ADD2(X, Y)
Used argument filtering: ADD2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, 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
            ↳ QDPAfsSolverProof

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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
Used argument filtering: SEL2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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, 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)
activate1(n__fib12(X1, X2)) -> fib12(X1, 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.