Termination w.r.t. Q proof of /tmp/tmpOPllK0/10.trs

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:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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:

DIV2(s(s(x))) → DIV2(x)
EVEN(s(s(x))) → EVEN(x)
COND1(true, x) → COND2(even(x), x)
NEQ(s(x), s(y)) → NEQ(x, y)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, x) → NEQ(x, 0)
COND1(true, x) → EVEN(x)
COND2(true, x) → DIV2(x)
COND2(false, x) → P(x)
COND2(true, x) → NEQ(x, 0)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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:

DIV2(s(s(x))) → DIV2(x)
EVEN(s(s(x))) → EVEN(x)
COND1(true, x) → COND2(even(x), x)
NEQ(s(x), s(y)) → NEQ(x, y)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, x) → NEQ(x, 0)
COND1(true, x) → EVEN(x)
COND2(true, x) → DIV2(x)
COND2(false, x) → P(x)
COND2(true, x) → NEQ(x, 0)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

DIV2(s(s(x))) → DIV2(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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.

DIV2(s(s(x))) → DIV2(x)

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

POL(DIV2(x₁)) = (2)x_1
POL(s(x₁)) = 1 + (2)x_1

The value of delta used in the strict ordering is 6.
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:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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:

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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.

EVEN(s(s(x))) → EVEN(x)

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

POL(EVEN(x₁)) = (2)x_1
POL(s(x₁)) = 1 + (2)x_1

The value of delta used in the strict ordering is 6.
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:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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

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

COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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.

COND2(false, x) → COND1(neq(x, 0), p(x))

The remaining pairs can at least be oriented weakly.

COND1(true, x) → COND2(even(x), x)
COND2(true, x) → COND1(neq(x, 0), div2(x))

Used ordering: Polynomial interpretation [25,35]:

POL(COND1(x₁, x₂)) = (1/2)x_2
POL(even(x₁)) = (1/4)x_1
POL(div2(x₁)) = (1/2)x_1
POL(COND2(x₁, x₂)) = (1/2)x_1 + (1/4)x_2
POL(true) = 0
POL(false) = 1
POL(p(x₁)) = (1/2)x_1
POL(s(x₁)) = 4 + (2)x_1
POL(0) = 0
POL(neq(x₁, x₂)) = (4)x_2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

even(0) → true
even(s(s(x))) → even(x)
even(s(0)) → false
div2(s(0)) → 0
div2(0) → 0
p(0) → 0
div2(s(s(x))) → s(div2(x))
p(s(x)) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

COND1(true, x) → COND2(even(x), x)
COND2(true, x) → COND1(neq(x, 0), div2(x))

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

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.

COND2(true, x) → COND1(neq(x, 0), div2(x))

The remaining pairs can at least be oriented weakly.

COND1(true, x) → COND2(even(x), x)

Used ordering: Polynomial interpretation [25,35]:

POL(COND1(x₁, x₂)) = 3 + (2)x_1 + (2)x_2
POL(even(x₁)) = 0
POL(div2(x₁)) = (1/4)x_1
POL(COND2(x₁, x₂)) = 4 + (3/2)x_2
POL(true) = 1/2
POL(false) = 1/4
POL(s(x₁)) = 1/2 + (4)x_1
POL(0) = 0
POL(neq(x₁, x₂)) = 1/4 + (1/2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

neq(s(x), 0) → true
div2(s(0)) → 0
div2(0) → 0
div2(s(s(x))) → s(div2(x))
neq(0, 0) → false

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

COND1(true, x) → COND2(even(x), x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.