Termination w.r.t. Q proof of /tmp/tmpCKzD7q/Ex4_7_77_Bor03

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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 + 2·x₂
POL(mark(x₁)) = x₁
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(tail(x₁)) = 2 + x₁
POL(top(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
PROPER(tail(X)) → TAIL(proper(X))
TAIL(mark(X)) → TAIL(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(zeros) → CONS(0, zeros)
PROPER(tail(X)) → PROPER(X)
TAIL(ok(X)) → TAIL(X)
PROPER(cons(X1, X2)) → PROPER(X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(tail(X)) → TAIL(active(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

CONS(mark(X1), X2) → CONS(X1, X2)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
PROPER(tail(X)) → TAIL(proper(X))
TAIL(mark(X)) → TAIL(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(zeros) → CONS(0, zeros)
PROPER(tail(X)) → PROPER(X)
TAIL(ok(X)) → TAIL(X)
PROPER(cons(X1, X2)) → PROPER(X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(tail(X)) → TAIL(active(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TAIL(ok(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(ok(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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(tail(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(tail(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)

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

PROPER(cons(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(tail(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(tail(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(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.

TOP(ok(X)) → TOP(active(X))

The remaining pairs can at least be oriented weakly.

TOP(mark(X)) → TOP(proper(X))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(mark(x₁)) = 1 + x₁
POL(ok(x₁)) = 1 + x₁
POL(proper(x₁)) = 1 + x₁
POL(tail(x₁)) = 1 + x₁
POL(zeros) = 1

The following usable rules [17] were oriented:

proper(tail(X)) → tail(proper(X))
proper(zeros) → ok(zeros)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ UsableRulesProof

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

TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))

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:

TOP(mark(X)) → TOP(proper(X))

Strictly oriented rules of the TRS R:

tail(mark(X)) → mark(tail(X))
cons(mark(X1), X2) → mark(cons(X1, X2))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(mark(x₁)) = 2 + x₁
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁
POL(tail(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
tail(ok(X)) → ok(tail(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.