Termination w.r.t. Q proof of /tmp/tmpyrwV0M/Ex1_Luc04b

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:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

head(cons(X, XS)) → X

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 2 + 2·x₁
POL(incr(x₁)) = x₁
POL(n__incr(x₁)) = x₁
POL(n__nats) = 0
POL(n__odds) = 0
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
tail(cons(X, XS)) → activate(XS)
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

tail(cons(X, XS)) → activate(XS)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(incr(x₁)) = x₁
POL(n__incr(x₁)) = x₁
POL(n__nats) = 0
POL(n__odds) = 0
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDS → PAIRS
ACTIVATE(n__nats) → NATS
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDS → PAIRS
ACTIVATE(n__nats) → NATS
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ODDS → INCR(pairs)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ODDS → INCR(pairs) at position [0] we obtained the following new rules:

ODDS → INCR(cons(0, n__incr(n__odds)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ACTIVATE(n__incr(X)) → INCR(activate(X)) at position [0] we obtained the following new rules:

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(n__nats)) → INCR(nats)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(n__nats)) → INCR(nats)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ACTIVATE(n__incr(n__nats)) → INCR(nats) at position [0] we obtained the following new rules:

ACTIVATE(n__incr(n__nats)) → INCR(n__nats)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__incr(n__nats)) → INCR(n__nats)
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ NonTerminationProof

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

The TRS P consists of the following rules:

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ODDS → INCR(cons(0, n__incr(n__odds)))
ACTIVATE(n__incr(n__odds)) → INCR(odds)
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

s = ACTIVATE(n__incr(n__nats)) evaluates to t =ACTIVATE(n__incr(n__nats))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats)))
with rule ACTIVATE(n__incr(n__nats)) → INCR(cons(0, n__incr(n__nats))) at position [] and matcher [ ]

INCR(cons(0, n__incr(n__nats))) → ACTIVATE(n__incr(n__nats))
with rule INCR(cons(X, XS)) → ACTIVATE(XS)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.