Termination w.r.t. Q proof of /tmp/tmpmMzBM_/parting02

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:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

Q is empty.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DOUBLE(s(x)) → DOUBLE(x)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
DOUBLELIST(cons(x, xs)) → DOUBLE(x)
DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
DOUBLELIST(cons(x, xs)) → DEL(first(cons(x, xs)), cons(x, xs))
DOUBLELIST(cons(x, xs)) → FIRST(cons(x, xs))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

DOUBLE(s(x)) → DOUBLE(x)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
DOUBLELIST(cons(x, xs)) → DOUBLE(x)
DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
DOUBLELIST(cons(x, xs)) → DEL(first(cons(x, xs)), cons(x, xs))
DOUBLELIST(cons(x, xs)) → FIRST(cons(x, xs))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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 [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

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

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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.[THIEMANN].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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.[THIEMANN].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF(false, x, y, xs) → DEL(x, xs)
The graph contains the following edges 2 >= 1, 4 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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.[THIEMANN].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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.[THIEMANN].

double(0)
double(s(x0))
doublelist(nil)
doublelist(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs))) at position [0] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs)) at position [0,0,0] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs)) at position [0,1] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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.[THIEMANN].

first(nil)
first(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Induction-Processor

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

This DP could be deleted by the Induction-Processor:
DOUBLELIST(cons(x', xs')) → DOUBLELIST(if(eq(x', x'), x', x', xs'))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(DOUBLELIST(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₂
POL(del(x₁, x₂)) = x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(if(x₁, x₂, x₃, x₄)) = 1 + x₄
POL(nil) = 0
POL(s(x₁)) = 0
POL(true) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(true, x110, y90, xs40) → xs40

The following formula is valid:
x':sort[a19],xs':sort[a4].if'(eq(x' , x' ), x' , x' , xs' )=true

The transformed set:
if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Induction-Processor

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ PisEmptyProof

                                                ↳ QTRS

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

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Induction-Processor

                                              ↳ AND

                                                ↳ QDP

                                                ↳ QTRS

                                                  ↳ QTRSRRRProof

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

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

Q is empty.

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

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

Q is empty.
Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

nil > false > true > eq₂
nil > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
0 > false > true > eq₂
0 > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
s₁ > eq₂
equal_bool₂ > eq₂
or₂ > eq₂
not₁ > false > true > eq₂
not₁ > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
isa_true₁ > false > true > eq₂
isa_true₁ > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
isa_false₁ > false > true > eq₂
isa_false₁ > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
equal_sort[a4]₂ > false > true > eq₂
equal_sort[a4]₂ > false > [if₄, del₂] > [cons₂, and₂] > [if'₄, del'₂] > eq₂
equal_sort[a19]₂ > eq₂
equal_sort[a37]₂ > true > eq₂
witness_sort[a37] > eq₂

Status:

if₄: [4,2,3,1]
eq₂: multiset
if'₄: [4,2,3,1]
equal_sort[a19]₂: multiset
true: multiset
or₂: multiset
and₂: [2,1]
0: multiset
del₂: [2,1]
del'₂: [2,1]
equal_bool₂: multiset
cons₂: [2,1]
not₁: [1]
equal_sort[a4]₂: multiset
witness_sort[a37]: multiset
isa_false₁: multiset
false: multiset
s₁: multiset
equal_sort[a37]₂: multiset
nil: multiset
isa_true₁: [1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ Induction-Processor

                                              ↳ AND

                                                ↳ QDP

                                                ↳ QTRS

                                                  ↳ QTRSRRRProof

                                                    ↳ QTRS

                                                      ↳ RisEmptyProof

                                                      ↳ RisEmptyProof

                                                      ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.