Termination w.r.t. Q proof of /tmp/tmpsZKZCf/5.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:

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

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) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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) → P(x)
COND(true, x) → EVEN(x)
EVEN(s(s(x))) → EVEN(x)
COND(true, x) → GR(x, 0)
COND(true, x) → AND(even(x), gr(x, 0))
COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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) → P(x)
COND(true, x) → EVEN(x)
EVEN(s(s(x))) → EVEN(x)
COND(true, x) → GR(x, 0)
COND(true, x) → AND(even(x), gr(x, 0))
COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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:

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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:

EVEN(s(s(x))) → EVEN(x)

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

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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:

EVEN(s(s(x))) → EVEN(x)

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:

EVEN(s(s(x))) → EVEN(x)
The graph contains the following edges 1 > 1

↳ 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) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
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)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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)

↳ 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) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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) → COND(and(even(x), gr(x, 0)), p(x))

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

POL(gr(x₁, x₂)) = 9/4 + (1/4)x_1
POL(even(x₁)) = 3/4
POL(true) = 3
POL(false) = 0
POL(p(x₁)) = (1/2)x_1
POL(s(x₁)) = 4 + (2)x_1
POL(and(x₁, x₂)) = 9/4 + (1/4)x_2
POL(COND(x₁, x₂)) = (1/4)x_1 + (1/4)x_2
POL(0) = 0

The value of delta used in the strict ordering is 3/64.
The following usable rules [17] were oriented:

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

↳ 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:

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

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
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.