Termination w.r.t. Q proof of /tmp/tmp7BU1k1/otto02.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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:

LENGTH(cons(x, y)) → LENGTH(y)
APP(x, y) → PLUS(length(x), length(y))
HELPA(c, l, ys, zs) → GE(c, l)
PLUS(x, s(y)) → PLUS(x, y)
HELPB(c, l, ys, zs) → TAKE(c, ys, zs)
TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)
HELPB(c, l, ys, zs) → HELPA(s(c), l, ys, zs)
APP(x, y) → LENGTH(x)
APP(x, y) → HELPA(0, plus(length(x), length(y)), x, y)
TAKE(s(c), cons(x, xs), ys) → TAKE(c, xs, ys)
HELPA(c, l, ys, zs) → IF(ge(c, l), c, l, ys, zs)
IF(false, c, l, ys, zs) → HELPB(c, l, ys, zs)
APP(x, y) → LENGTH(y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

LENGTH(cons(x, y)) → LENGTH(y)
APP(x, y) → PLUS(length(x), length(y))
HELPA(c, l, ys, zs) → GE(c, l)
PLUS(x, s(y)) → PLUS(x, y)
HELPB(c, l, ys, zs) → TAKE(c, ys, zs)
TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)
HELPB(c, l, ys, zs) → HELPA(s(c), l, ys, zs)
APP(x, y) → LENGTH(x)
APP(x, y) → HELPA(0, plus(length(x), length(y)), x, y)
TAKE(s(c), cons(x, xs), ys) → TAKE(c, xs, ys)
HELPA(c, l, ys, zs) → IF(ge(c, l), c, l, ys, zs)
IF(false, c, l, ys, zs) → HELPB(c, l, ys, zs)
APP(x, y) → LENGTH(y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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.

TAKE(s(c), nil, cons(y, ys)) → TAKE(c, nil, ys)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = (4)x_2
POL(TAKE(x₁, x₂, x₃)) = (4)x_1 + (3)x_3
POL(s(x₁)) = 1 + (4)x_1
POL(nil) = 0

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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.

GE(s(x), s(y)) → GE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(GE(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HELPA(c, l, ys, zs) → IF(ge(c, l), c, l, ys, zs)
IF(false, c, l, ys, zs) → HELPB(c, l, ys, zs)
HELPB(c, l, ys, zs) → HELPA(s(c), l, ys, zs)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

LENGTH(cons(x, y)) → LENGTH(y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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.

LENGTH(cons(x, y)) → LENGTH(y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(LENGTH(x₁)) = (4)x_1
POL(cons(x₁, x₂)) = 1 + (4)x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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.

PLUS(x, s(y)) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x₁, x₂)) = (4)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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