Termination w.r.t. Q proof of /tmp/tmpH_-XbV/ExIntrod_GM01

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:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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, L)) → X

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(activate(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 1 + 2·x₁
POL(incr(x₁)) = x₁
POL(n__adx(x₁)) = 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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:

adx(nil) → nil
nats → adx(zeros)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(activate(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(incr(x₁)) = x₁
POL(n__adx(x₁)) = 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nats) = 2
POL(nil) = 1
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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, L)) → activate(L)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(activate(x₁)) = x₁
POL(adx(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(incr(x₁)) = x₁
POL(n__adx(x₁)) = x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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, L)) → ACTIVATE(L)
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(X)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(X)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ADX(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ACTIVATE(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVATE(x₁)) = 2·x₁
POL(ADX(x₁)) = 2 + 2·x₁
POL(INCR(x₁)) = 2·x₁
POL(activate(x₁)) = x₁
POL(adx(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(n__adx(x₁)) = 1 + x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

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

ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

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

ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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__zeros)) → INCR(zeros) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( n__incr(x₁) ) =

/	0	\
\	0	/

/	1	1	\
\	1	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	1	\
\	1	0	/

x₂

M( n__adx(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( adx(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( incr(x₁) ) =

/	0	\
\	0	/

/	1	1	\
\	1	1	/

x₁

M( zeros ) =

/	1	\
\	1	/

M( n__zeros ) =

/	1	\
\	1	/

M( activate(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( 0 ) =

/	0	\
\	0	/

M( nil ) =

/	0	\
\	1	/

Tuple symbols:

M( ADX(x₁) ) =

[

]

x₁

M( INCR(x₁) ) =

[

]

x₁

M( ACTIVATE(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

incr(X) → n__incr(X)
zeros → cons(0, n__zeros)
zeros → n__zeros
adx(X) → n__adx(X)
incr(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
activate(n__adx(X)) → adx(activate(X))
activate(n__incr(X)) → incr(activate(X))
activate(X) → X
activate(n__zeros) → zeros

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

                                                    ↳ QDP

                                                      ↳ NonTerminationProof

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
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, L)) → ACTIVATE(L)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

s = ACTIVATE(n__adx(activate(n__zeros))) evaluates to t =ACTIVATE(n__adx(activate(n__zeros)))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

ACTIVATE(n__adx(activate(n__zeros))) → ACTIVATE(n__adx(n__zeros))
with rule activate(X) → X at position [0,0] and matcher [X / n__zeros]

ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
with rule ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros)) at position [] and matcher [ ]

ADX(cons(0, n__zeros)) → INCR(cons(0, n__adx(activate(n__zeros))))
with rule ADX(cons(X', L')) → INCR(cons(X', n__adx(activate(L')))) at position [] and matcher [X' / 0, L' / n__zeros]

INCR(cons(0, n__adx(activate(n__zeros)))) → ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(X, L)) → ACTIVATE(L)

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.