Termination w.r.t. Q proof of TRS/nontermin/CSREx8

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

  ↳ 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)

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)
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)

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)
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)

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)

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, XS)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SEL₂(x₁, x₂)) = 2·x₁ + x₂
POL(cons₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

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, 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.
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:

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)

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)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

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, 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.
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

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)

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