Termination w.r.t. Q proof of /tmp/tmpjZF4bv/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:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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:

a__nats → a__adx(a__zeros)
a__tail(cons(X, L)) → mark(L)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__adx(x₁)) = 2·x₁
POL(a__head(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 1
POL(a__tail(x₁)) = 1 + x₁
POL(a__zeros) = 0
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 1
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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:

a__adx(nil) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__adx(x₁)) = 2·x₁
POL(a__head(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__tail(x₁)) = x₁
POL(a__zeros) = 0
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 1
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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:

a__head(cons(X, L)) → mark(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__adx(x₁)) = x₁
POL(a__head(x₁)) = 1 + x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__tail(x₁)) = 2·x₁
POL(a__zeros) = 0
POL(adx(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(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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:

MARK(incr(X)) → A__INCR(mark(X))
MARK(tail(X)) → A__TAIL(mark(X))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(nats) → A__NATS
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(tail(X)) → A__TAIL(mark(X))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(nats) → A__NATS
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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 4 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

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

MARK(adx(X)) → A__ADX(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
MARK(adx(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__ADX(cons(X, L)) → MARK(X)
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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(adx(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
A__ADX(cons(X, L)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__ADX(x₁)) = 2 + 2·x₁
POL(A__INCR(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(a__adx(x₁)) = 1 + x₁
POL(a__head(x₁)) = 1 + 2·x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__tail(x₁)) = 1 + 2·x₁
POL(a__zeros) = 0
POL(adx(x₁)) = 1 + 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)) = 1 + 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(adx(X)) → A__ADX(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__ADX(x₁)) = 2 + 2·x₁
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = 2·x₁
POL(a__adx(x₁)) = 1 + x₁
POL(a__head(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__tail(x₁)) = x₁
POL(a__zeros) = 0
POL(adx(x₁)) = 1 + 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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

MARK(adx(X)) → A__ADX(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__INCR(cons(X, L)) → MARK(X)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__adx(x₁)) = 1 + x₁
POL(a__head(x₁)) = 0
POL(a__incr(x₁)) = 1 + x₁
POL(a__nats) = 0
POL(a__tail(x₁)) = 0
POL(a__zeros) = 0
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) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

mark(0) → 0
mark(s(X)) → s(mark(X))
a__adx(X) → adx(X)
a__incr(X) → incr(X)
a__zeros → zeros
a__nats → nats
a__tail(X) → tail(X)
a__head(X) → head(X)
mark(adx(X)) → a__adx(mark(X))
mark(incr(X)) → a__incr(mark(X))
mark(zeros) → a__zeros
mark(nats) → a__nats
mark(tail(X)) → a__tail(mark(X))
mark(head(X)) → a__head(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(nil) → nil
a__zeros → cons(0, zeros)
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__incr(nil) → nil

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ UsableRulesProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__zeros → cons(0, zeros)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ UsableRulesProof

                                            ↳ QDP

                                              ↳ QDPSizeChangeProof

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

MARK(s(X)) → MARK(X)
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 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:

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