Termination w.r.t. Q proof of /tmp/tmp8Xf7YA/Ex1_Luc04b

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__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
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__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
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__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
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__tail(cons(X, XS)) → mark(XS)
mark(nil) → nil
a__tail(X) → tail(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__head(x₁)) = 2·x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(a__tail(x₁)) = 2 + x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 0
POL(nil) = 2
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(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:

mark(tail(X)) → a__tail(mark(X))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__head(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(a__tail(x₁)) = 1 + 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·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)) = 2 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(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, XS)) → mark(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__head(x₁)) = 2 + x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(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(X) → head(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__head(x₁)) = 2 + 2·x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 1 + 2·x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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:

mark(head(X)) → a__head(mark(X))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__head(x₁)) = 2·x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 1 + 2·x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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 → nats

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 2
POL(a__odds) = 0
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 1
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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(pairs) → A__PAIRS
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__ODDS → A__PAIRS
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → A__NATS
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)
A__ODDS → A__INCR(a__pairs)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

MARK(pairs) → A__PAIRS
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__ODDS → A__PAIRS
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → A__NATS
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)
A__ODDS → A__INCR(a__pairs)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

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__ODDS → A__INCR(a__pairs)
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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(odds) → A__ODDS

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__INCR(x₁)) = 1 + x₁
POL(A__ODDS) = 2
POL(MARK(x₁)) = 1 + 2·x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__odds) = 1
POL(a__pairs) = 1
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 1
POL(pairs) = 1
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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, XS)) → MARK(X)
A__ODDS → A__INCR(a__pairs)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ 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, XS)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__incr(x₁)) = 1 + x₁
POL(a__nats) = 1
POL(a__odds) = 1
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = x₁
POL(incr(x₁)) = 1 + x₁
POL(mark(x₁)) = 1 + x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__odds → odds
a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ 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(cons(X1, X2)) → MARK(X1)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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)
A__INCR(cons(X, XS)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 1
POL(a__odds) = 1
POL(a__pairs) = 1
POL(cons(x₁, x₂)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 1
POL(odds) = 1
POL(pairs) = 1
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__odds → odds
a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(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

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ RRRPoloQTRSProof

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QDPSizeChangeProof

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

MARK(s(X)) → MARK(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:

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1