Termination w.r.t. Q proof of /tmp/tmpmIJCZs/double.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:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.

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

PERMUTE(false, x, b) → P(x)
ACK(s(x), 0) → ACK(x, s(0))
DOUBLE(x) → PERMUTE(x, x, a)
PERMUTE(x, y, a) → ISZERO(x)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PLUS(x, s(s(y))) → PLUS(s(x), y)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(0, x) → PLUS(x, s(0))
PLUS(s(x), y) → PLUS(x, s(y))
PERMUTE(false, x, b) → ACK(x, x)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

PERMUTE(false, x, b) → P(x)
ACK(s(x), 0) → ACK(x, s(0))
DOUBLE(x) → PERMUTE(x, x, a)
PERMUTE(x, y, a) → ISZERO(x)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PLUS(x, s(s(y))) → PLUS(s(x), y)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(0, x) → PLUS(x, s(0))
PLUS(s(x), y) → PLUS(x, s(y))
PERMUTE(false, x, b) → ACK(x, x)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

The set Q consists of the following terms:

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)

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

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(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.

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUS(x, s(s(y))) → PLUS(s(x), y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(PLUS(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 2 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

          ↳ QDP

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ QDP

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

The set Q consists of the following terms:

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(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.

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

          ↳ QDP

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

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:

ACK(s(x), 0) → ACK(x, s(0))
The graph contains the following edges 1 > 1
ACK(s(x), s(y)) → ACK(s(x), y)
The graph contains the following edges 1 >= 1, 2 > 2
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

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

PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

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

PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

The set Q consists of the following terms:

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)
p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(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.

double(x0)
permute(x0, x1, a)
permute(false, x0, b)
permute(true, x0, b)
permute(x0, x1, c)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(0, y1, a) → PERMUTE(true, 0, b)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(0, y1, a) → PERMUTE(true, 0, b)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)
isZero(0)
isZero(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.

isZero(0)
isZero(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c) we obtained the following new rules:

PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(0, x) → plus(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(0) → 0
p(s(x)) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(s(x)) → x
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c) at position [1] we obtained the following new rules:

PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), z0, c)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), z0, c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
p(s(x)) → x
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), z0, c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Narrowing

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), z0, c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule PERMUTE(false, s(z0), b) → PERMUTE(ack(s(z0), s(z0)), z0, c) at position [0] we obtained the following new rules:

PERMUTE(false, s(x0), b) → PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ ForwardInstantiation

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(false, s(x0), b) → PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule PERMUTE(y, x, c) → PERMUTE(x, y, a) we obtained the following new rules:

PERMUTE(x0, s(y_0), c) → PERMUTE(s(y_0), x0, a)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ ForwardInstantiation

                                                                  ↳ QDP

                                                                    ↳ QDPSizeChangeProof

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

PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
PERMUTE(x0, s(y_0), c) → PERMUTE(s(y_0), x0, a)
PERMUTE(false, s(x0), b) → PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → plus(x, s(0))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(0)) → s(x)
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, 0) → x

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
plus(x0, s(s(x1)))
plus(x0, s(0))
plus(x0, 0)

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:

PERMUTE(false, s(x0), b) → PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)
The graph contains the following edges 2 > 2
PERMUTE(x0, s(y_0), c) → PERMUTE(s(y_0), x0, a)
The graph contains the following edges 2 >= 1, 1 >= 2
PERMUTE(s(x0), y1, a) → PERMUTE(false, s(x0), b)
The graph contains the following edges 1 >= 2