Termination w.r.t. Q proof of /tmp/tmpGxBm

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:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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:

MAX(cons(u, l)) → MAX(l)
ST(n, l) → COND1(member(n, l), n, l)
GT(s(u), s(v)) → GT(u, v)
COND1(true, n, l) → ST(s(n), l)
COND2(false, n, l) → ST(s(n), l)
COND1(false, n, l) → MAX(l)
ST(n, l) → MEMBER(n, l)
COND1(false, n, l) → GT(n, max(l))
MEMBER(n, cons(m, l)) → MEMBER(n, l)
SORT(l) → ST(0, l)
EQUAL(s(x), s(y)) → EQUAL(x, y)
MAX(cons(u, l)) → IF(gt(u, max(l)), u, max(l))
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
MEMBER(n, cons(m, l)) → OR(equal(n, m), member(n, l))
MEMBER(n, cons(m, l)) → EQUAL(n, m)
MAX(cons(u, l)) → GT(u, max(l))

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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:

MAX(cons(u, l)) → MAX(l)
ST(n, l) → COND1(member(n, l), n, l)
GT(s(u), s(v)) → GT(u, v)
COND1(true, n, l) → ST(s(n), l)
COND2(false, n, l) → ST(s(n), l)
COND1(false, n, l) → MAX(l)
ST(n, l) → MEMBER(n, l)
COND1(false, n, l) → GT(n, max(l))
MEMBER(n, cons(m, l)) → MEMBER(n, l)
SORT(l) → ST(0, l)
EQUAL(s(x), s(y)) → EQUAL(x, y)
MAX(cons(u, l)) → IF(gt(u, max(l)), u, max(l))
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
MEMBER(n, cons(m, l)) → OR(equal(n, m), member(n, l))
MEMBER(n, cons(m, l)) → EQUAL(n, m)
MAX(cons(u, l)) → GT(u, max(l))

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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.

GT(s(u), s(v)) → GT(u, v)

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

POL(s(x₁)) = 1/2 + (13/4)x_1
POL(GT(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
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:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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:

MAX(cons(u, l)) → MAX(l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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.

MAX(cons(u, l)) → MAX(l)

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

POL(cons(x₁, x₂)) = 1/4 + (7/2)x_2
POL(MAX(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
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:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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

          ↳ QDP

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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.

EQUAL(s(x), s(y)) → EQUAL(x, y)

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

POL(EQUAL(x₁, x₂)) = (15/4)x_2
POL(s(x₁)) = 1/2 + (13/4)x_1

The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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

            ↳ QDPOrderProof

          ↳ QDP

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

MEMBER(n, cons(m, l)) → MEMBER(n, l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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.

MEMBER(n, cons(m, l)) → MEMBER(n, l)

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

POL(cons(x₁, x₂)) = 1/4 + (7/2)x_2
POL(MEMBER(x₁, x₂)) = (2)x_2

The value of delta used in the strict ordering is 1/2.
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:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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:

ST(n, l) → COND1(member(n, l), n, l)
COND1(false, n, l) → COND2(gt(n, max(l)), n, l)
COND1(true, n, l) → ST(s(n), l)
COND2(false, n, l) → ST(s(n), l)

The TRS R consists of the following rules:

sort(l) → st(0, l)
st(n, l) → cond1(member(n, l), n, l)
cond1(true, n, l) → cons(n, st(s(n), l))
cond1(false, n, l) → cond2(gt(n, max(l)), n, l)
cond2(true, n, l) → nil
cond2(false, n, l) → st(s(n), l)
member(n, nil) → false
member(n, cons(m, l)) → or(equal(n, m), member(n, l))
or(x, true) → true
or(x, false) → x
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
max(nil) → 0
max(cons(u, l)) → if(gt(u, max(l)), u, max(l))
if(true, u, v) → u
if(false, u, v) → v

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