Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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, 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, XS)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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, 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, XS)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))


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

FIB12(X, Y) -> ADD2(X, Y)
ADD2(s1(X), Y) -> ADD2(X, Y)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
FIB1(N) -> SEL2(N, fib12(s1(0), s1(0)))
FIB12(X, Y) -> FIB12(Y, add2(X, Y))
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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

FIB12(X, Y) -> ADD2(X, Y)
ADD2(s1(X), Y) -> ADD2(X, Y)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
FIB1(N) -> SEL2(N, fib12(s1(0), s1(0)))
FIB12(X, Y) -> FIB12(Y, add2(X, Y))
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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP

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

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

The TRS R consists of the following rules:

fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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, XS)
Used argument filtering: SEL2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ 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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

FIB12(X, Y) -> FIB12(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, 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, XS)

The set Q consists of the following terms:

fib1(x0)
fib12(x0, x1)
add2(0, x0)
add2(s1(x0), x1)
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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