Termination w.r.t. Q proof of /tmp/tmpwmPOQu/20.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, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

Q is empty.

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

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

The TRS R 2 is

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)

The signature Sigma is {cond1, cond2}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

COND2(false, x, y) → EQ(x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(true, x, y) → P(x)
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(true, x, y) → P(y)
EQ(s(x), s(y)) → EQ(x, y)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → AND(eq(x, y), gr(x, 0))
COND2(false, x, y) → GR(x, 0)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

COND2(false, x, y) → EQ(x, y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(true, x, y) → P(x)
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(true, x, y) → P(y)
EQ(s(x), s(y)) → EQ(x, y)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → AND(eq(x, y), gr(x, 0))
COND2(false, x, y) → GR(x, 0)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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)

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

              ↳ QDP

                ↳ UsableRulesProof

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

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND1(true, x, y) → COND2(gr(y, 0), x, y) at position [0] we obtained the following new rules:

COND1(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

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

COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y)) at position [0] we obtained the following new rules:

COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0)))
COND2(true, y0, 0) → COND2(false, p(y0), p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0)))
COND2(true, y0, 0) → COND2(false, p(y0), p(0))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(true, y0, s(x0)) → COND2(true, p(y0), p(s(x0))) at position [2] we obtained the following new rules:

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(true, y0, 0) → COND2(false, p(y0), p(0))
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(true, y0, 0) → COND2(false, p(y0), p(0)) at position [2] we obtained the following new rules:

COND2(true, y0, 0) → COND2(false, p(y0), 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

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

COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
COND2(true, y0, 0) → COND2(false, p(y0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y) we obtained the following new rules:

COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, 0) → COND2(false, p(y0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                ↳ QDP

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

COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                ↳ QDP

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

COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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.

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                ↳ QDP

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

COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND2(false, z0, 0) → COND1(and(eq(z0, 0), gr(z0, 0)), z0, 0) at position [0] we obtained the following new rules:

COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                ↳ QDP

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

COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0)
COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(false, s(x0), 0) → COND1(and(eq(s(x0), 0), true), s(x0), 0) at position [0,0] we obtained the following new rules:

COND2(false, s(x0), 0) → COND1(and(false, true), s(x0), 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                ↳ QDP

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

COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND2(false, s(x0), 0) → COND1(and(false, true), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                ↳ QDP

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

COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                ↳ QDP

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

COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(false, s(x0), 0) → COND1(and(false, gr(s(x0), 0)), s(x0), 0) at position [0] we obtained the following new rules:

COND2(false, s(x0), 0) → COND1(false, s(x0), 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, s(x0), 0) → COND1(false, s(x0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

gr(0, x) → false
and(true, true) → true
and(x, false) → false
eq(0, 0) → true
and(false, x) → false

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(false, 0, 0) → COND1(and(true, gr(0, 0)), 0, 0) at position [0,1] we obtained the following new rules:

COND2(false, 0, 0) → COND1(and(true, false), 0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ DependencyGraphProof

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)
COND2(false, 0, 0) → COND1(and(true, false), 0, 0)

The TRS R consists of the following rules:

gr(0, x) → false
and(true, true) → true
and(x, false) → false
eq(0, 0) → true
and(false, x) → false

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ DependencyGraphProof

                                                                                            ↳ QDP

                                                                                              ↳ UsableRulesProof

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

gr(0, x) → false
and(true, true) → true
and(x, false) → false
eq(0, 0) → true
and(false, x) → false

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ DependencyGraphProof

                                                                                            ↳ QDP

                                                                                              ↳ UsableRulesProof

                                                                                                ↳ QDP

                                                                                                  ↳ QReductionProof

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

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

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.

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ DependencyGraphProof

                                                                                            ↳ QDP

                                                                                              ↳ UsableRulesProof

                                                                                                ↳ QDP

                                                                                                  ↳ QReductionProof

                                                                                                    ↳ QDP

                                                                                                      ↳ Rewriting

                                                ↳ QDP

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

COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(false, 0, 0) → COND1(and(eq(0, 0), false), 0, 0) at position [0] we obtained the following new rules:

COND2(false, 0, 0) → COND1(false, 0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ DependencyGraphProof

                                                                    ↳ QDP

                                                                      ↳ UsableRulesProof

                                                                        ↳ QDP

                                                                          ↳ Rewriting

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ DependencyGraphProof

                                                                                            ↳ QDP

                                                                                              ↳ UsableRulesProof

                                                                                                ↳ QDP

                                                                                                  ↳ QReductionProof

                                                                                                    ↳ QDP

                                                                                                      ↳ Rewriting

                                                                                                        ↳ QDP

                                                                                                          ↳ DependencyGraphProof

                                                ↳ QDP

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

COND2(false, 0, 0) → COND1(false, 0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

The TRS R consists of the following rules:

eq(0, 0) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                ↳ QDP

                                                  ↳ UsableRulesProof

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

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

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.

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ AND

                                                ↳ QDP

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ QDPSizeChangeProof

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

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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:

COND2(true, y0, s(x0)) → COND2(true, p(y0), x0)
The graph contains the following edges 1 >= 1, 3 > 3