Termination w.r.t. Q proof of /tmp/tmpJaDoig/ExIntrod_GM01

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

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

active(nats) → mark(adx(zeros))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 1
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:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

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

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(head(x₁)) = 2 + 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 2
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → INCR(mark(X))
TAIL(active(X)) → TAIL(X)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(nil)) → MARK(nil)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
INCR(active(X)) → INCR(X)
HEAD(mark(X)) → HEAD(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(adx(cons(X, L))) → ADX(L)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → CONS(0, zeros)
ADX(mark(X)) → ADX(X)
ACTIVE(incr(cons(X, L))) → S(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
MARK(head(X)) → HEAD(mark(X))
ADX(active(X)) → ADX(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
INCR(mark(X)) → INCR(X)
MARK(tail(X)) → TAIL(mark(X))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
MARK(s(X)) → S(mark(X))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
MARK(adx(X)) → ADX(mark(X))
CONS(active(X1), X2) → CONS(X1, X2)
MARK(nats) → ACTIVE(nats)
MARK(nil) → ACTIVE(nil)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → INCR(mark(X))
TAIL(active(X)) → TAIL(X)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(nil)) → MARK(nil)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
INCR(active(X)) → INCR(X)
HEAD(mark(X)) → HEAD(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(adx(cons(X, L))) → ADX(L)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → CONS(0, zeros)
ADX(mark(X)) → ADX(X)
ACTIVE(incr(cons(X, L))) → S(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
MARK(head(X)) → HEAD(mark(X))
ADX(active(X)) → ADX(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
INCR(mark(X)) → INCR(X)
MARK(tail(X)) → TAIL(mark(X))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
MARK(s(X)) → S(mark(X))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
MARK(adx(X)) → ADX(mark(X))
CONS(active(X1), X2) → CONS(X1, X2)
MARK(nats) → ACTIVE(nats)
MARK(nil) → ACTIVE(nil)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 7 SCCs with 17 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(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:

TAIL(active(X)) → TAIL(X)
The graph contains the following edges 1 > 1
TAIL(mark(X)) → TAIL(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)

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

HEAD(mark(X)) → HEAD(X)
The graph contains the following edges 1 > 1
HEAD(active(X)) → HEAD(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ADX(active(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ADX(active(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

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

ADX(active(X)) → ADX(X)
The graph contains the following edges 1 > 1
ADX(mark(X)) → ADX(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
S(active(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

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

CONS(X1, active(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

                  ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

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

CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

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

INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

                  ↳ QDP

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

INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)

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

INCR(mark(X)) → INCR(X)
The graph contains the following edges 1 > 1
INCR(active(X)) → INCR(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(adx(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

MARK(tail(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(active(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2 + x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

MARK(head(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(adx(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 1 + 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

MARK(adx(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(active(x₁)) = x₁
POL(adx(x₁)) = 1 + 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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.

ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))

The remaining pairs can at least be oriented weakly.

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(adx(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁
POL(head(x₁)) = 0
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(zeros) → mark(cons(0, zeros))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
mark(zeros) → active(zeros)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
active(incr(nil)) → mark(nil)
mark(nats) → active(nats)
mark(0) → active(0)
mark(nil) → active(nil)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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.

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(adx(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁
POL(head(x₁)) = 0
POL(incr(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 1
POL(s(x₁)) = x₁
POL(tail(x₁)) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(zeros) → mark(cons(0, zeros))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
mark(zeros) → active(zeros)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
active(incr(nil)) → mark(nil)
mark(nats) → active(nats)
mark(0) → active(0)
mark(nil) → active(nil)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

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

MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

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

MARK(s(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

MARK(s(X)) → MARK(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                    ↳ QDPOrderProof

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

MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

R is empty.
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.

ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

MARK(zeros) → ACTIVE(zeros)

Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1 + (2)x_1
POL(MARK(x₁)) = x_1
POL(zeros) = 4
POL(0) = 1
POL(ACTIVE(x₁)) = 4

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                    ↳ QDPOrderProof

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

MARK(zeros) → ACTIVE(zeros)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVE(zeros) → MARK(cons(0, zeros))

The remaining pairs can at least be oriented weakly.

MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
zeros = zeros
ACTIVE(x1) = ACTIVE(x1)
cons(x1, x2) = x1
0 = 0

Recursive path order with status [2].
Quasi-Precedence:

[MARK₁, ACTIVE₁] > zeros > 0

Status:

MARK₁: [1]
zeros: multiset
0: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

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

MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → MARK(X1)

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