Termination w.r.t. Q proof of /tmp/tmpZNkQDl/ExIntrod

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), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

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

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

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

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), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

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

adx(nil) → nil
head(cons(X, L)) → X
tail(cons(X, L)) → L

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2 + x₁
POL(incr(x₁)) = x₁
POL(nats) = 0
POL(nil) = 2
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), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

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), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

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

nats → adx(zeros)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(incr(x₁)) = x₁
POL(nats) = 2
POL(nil) = 0
POL(s(x₁)) = x₁
POL(zeros) = 0

↳ 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), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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:

ZEROS → ZEROS
ADX(cons(X, L)) → ADX(L)
ADX(cons(X, L)) → INCR(cons(X, adx(L)))
INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

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:

ZEROS → ZEROS
ADX(cons(X, L)) → ADX(L)
ADX(cons(X, L)) → INCR(cons(X, adx(L)))
INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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 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:

ZEROS → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                  ↳ QDP

                  ↳ QDP

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

ZEROS → ZEROS

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

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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.

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ NonTerminationProof

                  ↳ QDP

                  ↳ QDP

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

ZEROS → ZEROS

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:

ZEROS → ZEROS

The TRS R consists of the following rules:none

s = ZEROS evaluates to t =ZEROS

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 ZEROS to ZEROS.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

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

INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                  ↳ QDP

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

INCR(cons(X, L)) → INCR(L)

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

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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.

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

                  ↳ QDP

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

INCR(cons(X, L)) → INCR(L)

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:

INCR(cons(X, L)) → INCR(L)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

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

ADX(cons(X, L)) → ADX(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

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

ADX(cons(X, L)) → ADX(L)

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

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(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.

adx(nil)
tail(cons(x0, x1))
zeros
adx(cons(x0, x1))
incr(nil)
head(cons(x0, x1))
nats
incr(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

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

ADX(cons(X, L)) → ADX(L)

From the DPs we obtained the following set of size-change graphs:

ADX(cons(X, L)) → ADX(L)
The graph contains the following edges 1 > 1