Termination w.r.t. Q proof of /tmp/tmpSNsjPq/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(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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)) = 2·x₁ + 2·x₂
POL(head(x₁)) = 2 + 2·x₁
POL(incr(x₁)) = x₁
POL(n__incr(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

nats → cons(0, n__incr(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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(nats))
pairs → cons(0, n__incr(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)
activate(n__incr(X)) → incr(X)
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)) = 2·x₁ + 2·x₂
POL(incr(x₁)) = 2·x₁
POL(n__incr(x₁)) = 2·x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = 2·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(nats))
pairs → cons(0, n__incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
activate(n__incr(X)) → incr(X)
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)
ACTIVATE(n__incr(X)) → INCR(X)
ODDS → INCR(pairs)
NATS → NATS
PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:

nats → cons(0, n__incr(nats))
pairs → cons(0, n__incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
activate(n__incr(X)) → incr(X)
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)
ACTIVATE(n__incr(X)) → INCR(X)
ODDS → INCR(pairs)
NATS → NATS
PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:

nats → cons(0, n__incr(nats))
pairs → cons(0, n__incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
activate(n__incr(X)) → incr(X)
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 3 SCCs with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

                  ↳ QDP

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)

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:

ACTIVATE(n__incr(X)) → INCR(X)
The graph contains the following edges 1 > 1
INCR(cons(X, XS)) → ACTIVATE(XS)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

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

PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ NonTerminationProof

                  ↳ QDP

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

PAIRS → ODDS
ODDS → PAIRS

R is empty.
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:

PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:none

s = ODDS evaluates to t =ODDS

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

ODDS → PAIRS
with rule ODDS → PAIRS at position [] and matcher [ ]

PAIRS → ODDS
with rule PAIRS → ODDS

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

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

NATS → NATS

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ NonTerminationProof

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

NATS → NATS

R is empty.
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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

NATS → NATS

The TRS R consists of the following rules:none

s = NATS evaluates to t =NATS

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS to NATS.