Termination w.r.t. Q proof of /tmp/tmp6qScZg/Ex8

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

↳ 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, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

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

FIB1(X, Y) → ADD(X, Y)
FIB(N) → FIB1(s(0), s(0))
FIB1(X, Y) → FIB1(Y, add(X, Y))
SEL(s(N), cons(X, XS)) → SEL(N, XS)
ADD(s(X), Y) → ADD(X, Y)
FIB(N) → SEL(N, fib1(s(0), s(0)))

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

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:

FIB1(X, Y) → ADD(X, Y)
FIB(N) → FIB1(s(0), s(0))
FIB1(X, Y) → FIB1(Y, add(X, Y))
SEL(s(N), cons(X, XS)) → SEL(N, XS)
ADD(s(X), Y) → ADD(X, Y)
FIB(N) → SEL(N, fib1(s(0), s(0)))

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

R is empty.
The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

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:

SEL(s(N), cons(X, XS)) → SEL(N, XS)
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:

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, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

R is empty.
The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

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

          ↳ QDP

            ↳ UsableRulesProof

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

FIB1(X, Y) → FIB1(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, fib1(Y, 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, XS)

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

FIB1(X, Y) → FIB1(Y, add(X, Y))

The TRS R consists of the following rules:

add(0, X) → X
add(s(X), Y) → s(add(X, Y))

The set Q consists of the following terms:

add(0, x0)
fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
add(s(x0), x1)
sel(0, cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

fib(x0)
sel(s(x0), cons(x1, x2))
fib1(x0, x1)
sel(0, cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ MNOCProof

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

FIB1(X, Y) → FIB1(Y, add(X, Y))

The TRS R consists of the following rules:

add(0, X) → X
add(s(X), Y) → s(add(X, Y))

The set Q consists of the following terms:

add(0, x0)
add(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ NonTerminationProof

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

FIB1(X, Y) → FIB1(Y, add(X, Y))

The TRS R consists of the following rules:

add(0, X) → X
add(s(X), Y) → s(add(X, Y))

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

FIB1(X, Y) → FIB1(Y, add(X, Y))

The TRS R consists of the following rules:

add(0, X) → X
add(s(X), Y) → s(add(X, Y))

s = FIB1(X, Y) evaluates to t =FIB1(Y, add(X, Y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [Y / add(X, Y), X / Y]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FIB1(X, Y) to FIB1(Y, add(X, Y)).