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

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:

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The signature Sigma is {cond}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

COND(true, x, y, z) → GR(y, z)
COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)
COND(true, x, y, z) → P(y)
COND(true, x, y, z) → P(x)
COND(true, x, y, z) → GR(x, z)
COND(true, x, y, z) → AND(gr(x, z), gr(y, z))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

COND(true, x, y, z) → GR(y, z)
COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)
COND(true, x, y, z) → P(y)
COND(true, x, y, z) → P(x)
COND(true, x, y, z) → GR(x, z)
COND(true, x, y, z) → AND(gr(x, z), gr(y, z))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

R is empty.
The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

GR(s(x), s(y)) → GR(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GR(s(x), s(y)) → GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond(true, x0, x1, x2)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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.

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

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

POL(COND(x₁, x₂, x₃, x₄)) = (1/2)x_1 + (5/4)x_3
POL(gr(x₁, x₂)) = (1/4)x_1
POL(true) = 1/4
POL(false) = 0
POL(p(x₁)) = (1/4)x_1
POL(s(x₁)) = 1 + (4)x_1
POL(and(x₁, x₂)) = (5/2)x_2
POL(0) = 0

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

p(s(x)) → x
gr(0, x) → false
gr(0, 0) → false
gr(s(x), s(y)) → gr(x, y)
gr(s(x), 0) → true
and(x, false) → false
and(true, true) → true
p(0) → 0
and(false, x) → false

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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