Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(head(cons(X, XS))) → mark(X)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(tail(cons(X, XS))) → mark(XS)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
ACTIVE(nats) → CONS(0, incr(nats))
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)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
INCR(active(X)) → INCR(X)
HEAD(mark(X)) → HEAD(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(pairs) → ACTIVE(pairs)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(nats) → INCR(nats)
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(odds) → INCR(pairs)
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
MARK(head(X)) → HEAD(mark(X))
S(active(X)) → S(X)
S(mark(X)) → S(X)
INCR(mark(X)) → INCR(X)
MARK(tail(X)) → TAIL(mark(X))
MARK(s(X)) → S(mark(X))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(pairs) → INCR(odds)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
ACTIVE(pairs) → CONS(0, incr(odds))
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, XS))) → CONS(s(X), incr(XS))
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(0) → ACTIVE(0)
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
ACTIVE(nats) → CONS(0, incr(nats))
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)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
INCR(active(X)) → INCR(X)
HEAD(mark(X)) → HEAD(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(pairs) → ACTIVE(pairs)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(nats) → INCR(nats)
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(odds) → INCR(pairs)
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
MARK(head(X)) → HEAD(mark(X))
S(active(X)) → S(X)
S(mark(X)) → S(X)
INCR(mark(X)) → INCR(X)
MARK(tail(X)) → TAIL(mark(X))
MARK(s(X)) → S(mark(X))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(pairs) → INCR(odds)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
ACTIVE(pairs) → CONS(0, incr(odds))
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, XS))) → CONS(s(X), incr(XS))
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(0) → ACTIVE(0)
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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 6 SCCs with 14 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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

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

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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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

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

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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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

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

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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

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

                    ↳ UsableRulesProof

                  ↳ 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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

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

                    ↳ RuleRemovalProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
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(head(X)) → ACTIVE(head(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
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(head(X)) → ACTIVE(head(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(head(x₁)) = 0
POL(incr(x₁)) = 1
POL(mark(x₁)) = 0
POL(nats) = 1
POL(odds) = 1
POL(pairs) = 1
POL(s(x₁)) = 0
POL(tail(x₁)) = 0

The following usable rules [17] were oriented:

tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(s(X)) → MARK(X)
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(pairs) → MARK(cons(0, incr(odds)))
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(nats) → MARK(cons(0, incr(nats)))

The remaining pairs can at least be oriented weakly.

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(pairs) → ACTIVE(pairs)
MARK(s(X)) → MARK(X)
MARK(odds) → ACTIVE(odds)
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
incr(x1) = x1
cons(x1, x2) = x1
MARK(x1) = MARK(x1)
s(x1) = x1
pairs = pairs
0 = 0
odds = odds
mark(x1) = x1
nats = nats
tail(x1) = tail
active(x1) = x1
head(x1) = head(x1)

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

odds > pairs > [ACTIVE₁, MARK₁] > nats > 0

Status:

tail: multiset
odds: multiset
MARK₁: [1]
head₁: multiset
pairs: multiset
nats: multiset
0: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented:

tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(0) → active(0)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(pairs) → ACTIVE(pairs)
MARK(s(X)) → MARK(X)
MARK(odds) → ACTIVE(odds)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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 3 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
incr(x1) = x1
cons(x1, x2) = cons(x1)
MARK(x1) = MARK(x1)
s(x1) = x1
mark(x1) = x1
tail(x1) = tail
active(x1) = x1
head(x1) = head
0 = 0
nats = nats
pairs = pairs
odds = odds

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

[ACTIVE₁, MARK₁] > cons₁
tail > cons₁
head > cons₁
nats > 0 > cons₁
[pairs, odds] > 0 > cons₁

Status:

tail: multiset
odds: multiset
cons₁: [1]
MARK₁: [1]
head: multiset
pairs: multiset
nats: multiset
0: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented:

tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
incr(x1) = x1
cons(x1, x2) = x2
MARK(x1) = MARK(x1)
s(x1) = s(x1)
mark(x1) = x1
tail(x1) = tail
active(x1) = x1
head(x1) = x1
0 = 0
nats = nats
pairs = pairs
odds = odds

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

[ACTIVE₁, MARK₁] > s₁
[pairs, odds]

Status:

tail: multiset
odds: multiset
MARK₁: [1]
pairs: multiset
s₁: [1]
nats: multiset
0: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented:

tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.

MARK(incr(X)) → MARK(X)
MARK(incr(X)) → ACTIVE(incr(mark(X)))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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 0 SCCs with 1 less node.