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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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, 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
            ↳ 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, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(SEL2(x1, x2)) = x1   
POL(activate1(x1)) = x1   
POL(add2(x1, x2)) = 0   
POL(cons2(x1, x2)) = 1   
POL(fib12(x1, x2)) = 1 + x1 + x2   
POL(n__fib12(x1, x2)) = 1 + x1 + x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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, 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.