Termination w.r.t. Q proof of /tmp/tmpwuDM76/otto10.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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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)
HELP(c, l, cons(x, y), z) → GE(c, l)
REV(x) → LENGTH(x)
REV(x) → IF(x, eq(0, length(x)), nil, 0, length(x))
HELP(c, l, cons(x, y), z) → APPEND(y, cons(x, nil))
IF(x, false, z, c, l) → HELP(s(c), l, x, z)
GE(s(x), s(y)) → GE(x, y)
APPEND(cons(x, y), z) → APPEND(y, z)
HELP(c, l, cons(x, y), z) → IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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)
HELP(c, l, cons(x, y), z) → GE(c, l)
REV(x) → LENGTH(x)
REV(x) → IF(x, eq(0, length(x)), nil, 0, length(x))
HELP(c, l, cons(x, y), z) → APPEND(y, cons(x, nil))
IF(x, false, z, c, l) → HELP(s(c), l, x, z)
GE(s(x), s(y)) → GE(x, y)
APPEND(cons(x, y), z) → APPEND(y, z)
HELP(c, l, cons(x, y), z) → IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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)) = 2 + (4)x_2

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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

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

APPEND(cons(x, y), z) → APPEND(y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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.

APPEND(cons(x, y), z) → APPEND(y, z)

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

POL(cons(x₁, x₂)) = 2 + (4)x_2
POL(APPEND(x₁, x₂)) = (4)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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

            ↳ QDPOrderProof

          ↳ 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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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)) = (1/2)x_1
POL(s(x₁)) = 1/2 + (4)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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

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

IF(x, false, z, c, l) → HELP(s(c), l, x, z)
HELP(c, l, cons(x, y), z) → IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
rev(x) → if(x, eq(0, length(x)), nil, 0, length(x))
if(x, true, z, c, l) → z
if(x, false, z, c, l) → help(s(c), l, x, z)
help(c, l, cons(x, y), z) → if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
append(nil, y) → y
append(cons(x, y), z) → cons(x, append(y, z))
length(nil) → 0
length(cons(x, y)) → s(length(y))

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