Termination w.r.t. Q proof of /tmp/tmpjEsDcE/cade13t.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:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

DIV(x, s(y)) → D(x, s(y), 0)
COND(true, x, y, z) → PLUS(s(y), z)
PLUS(n, s(m)) → PLUS(n, m)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)
D(x, s(y), z) → GE(x, z)
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

DIV(x, s(y)) → D(x, s(y), 0)
COND(true, x, y, z) → PLUS(s(y), z)
PLUS(n, s(m)) → PLUS(n, m)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)
D(x, s(y), z) → GE(x, z)
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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(n, s(m)) → PLUS(n, m)

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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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

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

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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(u), s(v)) → GE(u, v)

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

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

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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

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

D(x, s(y), z) → COND(ge(x, z), x, y, z)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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