Termination w.r.t. Q proof of /tmp/tmprzzGf1/LPAR

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:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.

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

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

int(s(x), 0) → nil
int(x, x) → cons(x, nil)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(int(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(intlist(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

intlist(nil) → nil
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.

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

intlist(nil) → nil
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

intlist(nil) → nil
intlist(cons(x, nil)) → cons(s(x), nil)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(int(x₁, x₂)) = 2·x₁ + x₂
POL(intlist(x₁)) = 2·x₁
POL(nil) = 2
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

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

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

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

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)
INT(s(x), s(y)) → INTLIST(int(x, y))
INTLIST(cons(x, y)) → INTLIST(y)

The TRS R consists of the following rules:

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)
INT(s(x), s(y)) → INTLIST(int(x, y))
INTLIST(cons(x, y)) → INTLIST(y)

The TRS R consists of the following rules:

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

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

INTLIST(cons(x, y)) → INTLIST(y)

The TRS R consists of the following rules:

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

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

INTLIST(cons(x, y)) → INTLIST(y)

R is empty.
The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

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

INTLIST(cons(x, y)) → INTLIST(y)

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:

INTLIST(cons(x, y)) → INTLIST(y)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

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

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)

The TRS R consists of the following rules:

intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))

The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

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

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)

R is empty.
The set Q consists of the following terms:

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

intlist(cons(x0, x1))
int(s(x0), s(x1))
int(0, s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ Overlay + Local Confluence

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

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

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT(x, y)

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

INT(s(x), s(y)) → INT(x, y)
The graph contains the following edges 1 > 1, 2 > 2
INT(0, s(y)) → INT(s(0), s(y))
The graph contains the following edges 2 >= 2