Termination w.r.t. Q proof of /tmp/tmpKLnJ7s/tower

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:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

tower(x) → f(a, x, s(0))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a) = 0
POL(b) = 0
POL(double(x₁)) = x₁
POL(exp(x₁)) = x₁
POL(f(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(half(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tower(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

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

f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f(a, 0, y) → y

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a) = 1
POL(b) = 1
POL(double(x₁)) = x₁
POL(exp(x₁)) = x₁
POL(f(x₁, x₂, x₃)) = 2 + x₁ + 2·x₂ + 2·x₃
POL(half(x₁)) = x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

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

exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The TRS R 2 is

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))

The signature Sigma is {f}

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

HALF(s(0)) → HALF(0)
F(a, s(x), y) → F(b, y, s(x))
HALF(s(s(x))) → HALF(x)
DOUBLE(s(x)) → DOUBLE(x)
EXP(s(x)) → EXP(x)
EXP(s(x)) → DOUBLE(exp(x))
HALF(0) → DOUBLE(0)
F(b, y, x) → EXP(y)
F(b, y, x) → HALF(x)
F(b, y, x) → F(a, half(x), exp(y))

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

HALF(s(0)) → HALF(0)
F(a, s(x), y) → F(b, y, s(x))
HALF(s(s(x))) → HALF(x)
DOUBLE(s(x)) → DOUBLE(x)
EXP(s(x)) → EXP(x)
EXP(s(x)) → DOUBLE(exp(x))
HALF(0) → DOUBLE(0)
F(b, y, x) → EXP(y)
F(b, y, x) → HALF(x)
F(b, y, x) → F(a, half(x), exp(y))

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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.

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

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

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

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

HALF(s(s(x))) → HALF(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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.

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

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

HALF(s(s(x))) → HALF(x)

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

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

EXP(s(x)) → EXP(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

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

EXP(s(x)) → EXP(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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.

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

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

EXP(s(x)) → EXP(x)

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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.

f(a, s(x0), x1)
f(b, x0, x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(b, y, x) → F(a, half(x), exp(y)) at position [1] we obtained the following new rules:

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(b, y0, s(0)) → F(a, half(0), exp(y0)) at position [1] we obtained the following new rules:

F(b, y0, s(0)) → F(a, double(0), exp(y0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(0)) → F(a, double(0), exp(y0))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(b, y0, s(0)) → F(a, double(0), exp(y0)) at position [1] we obtained the following new rules:

F(b, y0, s(0)) → F(a, 0, exp(y0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, 0, exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                        ↳ SemLabProof

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

F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                          ↳ QDP

                                                        ↳ SemLabProof

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

F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

Q is empty.
We have to consider all (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 0
half: 0
b: 1
s: 0
double: 0
0: 0
F: 0
exp: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1-1-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.1(y0))
F.0-0-1(a., s.0(s.0(y_1)), x1) → F.1-1-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.0(y0))
F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-1-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.1(y0))
F.0-0-1(a., s.0(s.1(y_1)), x1) → F.1-1-0(b., x1, s.0(s.1(y_1)))
F.0-0-0(a., s.0(s.1(y_1)), x1) → F.1-0-0(b., x1, s.0(s.1(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

double.0(s.1(x)) → s.0(s.0(double.1(x)))
half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                        ↳ SemLabProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

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

F.1-1-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.1(y0))
F.0-0-1(a., s.0(s.0(y_1)), x1) → F.1-1-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.0(y0))
F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-1-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.1(y0))
F.0-0-1(a., s.0(s.1(y_1)), x1) → F.1-1-0(b., x1, s.0(s.1(y_1)))
F.0-0-0(a., s.0(s.1(y_1)), x1) → F.1-0-0(b., x1, s.0(s.1(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

double.0(s.1(x)) → s.0(s.0(double.1(x)))
half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                        ↳ SemLabProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesReductionPairsProof

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

double.0(s.1(x)) → s.0(s.0(double.1(x)))
half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
double.0(s.1(x)) → s.0(s.0(double.1(x)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0.) = 0
POL(F.0-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(F.1-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(a.) = 0
POL(b.) = 0
POL(double.0(x₁)) = x₁
POL(double.1(x₁)) = x₁
POL(exp.0(x₁)) = x₁
POL(exp.1(x₁)) = x₁
POL(half.0(x₁)) = x₁
POL(half.1(x₁)) = x₁
POL(s.0(x₁)) = x₁
POL(s.1(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                        ↳ SemLabProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesReductionPairsProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The remaining pairs can at least be oriented weakly.

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))

Used ordering: Polynomial interpretation [25]:

POL(0.) = 0
POL(F.0-0-0(x₁, x₂, x₃)) = x₂
POL(F.1-0-0(x₁, x₂, x₃)) = x₃
POL(a.) = 0
POL(b.) = 0
POL(double.0(x₁)) = 0
POL(exp.0(x₁)) = 0
POL(half.0(x₁)) = x₁
POL(s.0(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ AAECC Innermost

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                        ↳ SemLabProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesReductionPairsProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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