Termination w.r.t. Q proof of /tmp/tmpwoOpNd/aprove1.srs

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:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

The set Q is empty.
We have obtained the following QTRS:

0(p(x)) → p(s(s(0(x))))
s(p(x)) → x
s(p(p(x))) → p(x)
s(f(x)) → s(g(x))
g(x) → half(s(i(x)))
i(x) → p(f(x))
0(half(x)) → half(s(s(0(x))))
s(s(half(x))) → s(s(p(p(half(s(x))))))
0(x) → x
0(rd(x)) → rd(0(0(0(0(0(0(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

0(p(x)) → p(s(s(0(x))))
s(p(x)) → x
s(p(p(x))) → p(x)
s(f(x)) → s(g(x))
g(x) → half(s(i(x)))
i(x) → p(f(x))
0(half(x)) → half(s(s(0(x))))
s(s(half(x))) → s(s(p(p(half(s(x))))))
0(x) → x
0(rd(x)) → rd(0(0(0(0(0(0(x)))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

The set Q is empty.
We have obtained the following QTRS:

0(p(x)) → p(s(s(0(x))))
s(p(x)) → x
s(p(p(x))) → p(x)
s(f(x)) → s(g(x))
g(x) → half(s(i(x)))
i(x) → p(f(x))
0(half(x)) → half(s(s(0(x))))
s(s(half(x))) → s(s(p(p(half(s(x))))))
0(x) → x
0(rd(x)) → rd(0(0(0(0(0(0(x)))))))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

0(p(x)) → p(s(s(0(x))))
s(p(x)) → x
s(p(p(x))) → p(x)
s(f(x)) → s(g(x))
g(x) → half(s(i(x)))
i(x) → p(f(x))
0(half(x)) → half(s(s(0(x))))
s(s(half(x))) → s(s(p(p(half(s(x))))))
0(x) → x
0(rd(x)) → rd(0(0(0(0(0(0(x)))))))

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:

F(s(x1)) → G(s(x1))
G(x1) → HALF(x1)
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(s(s(x1))) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
HALF(0(x1)) → 0¹(s(s(half(x1))))
RD(0(x1)) → RD(x1)
RD(0(x1)) → 0¹(0(0(rd(x1))))
P(0(x1)) → P(x1)
HALF(0(x1)) → HALF(x1)
RD(0(x1)) → 0¹(0(0(0(0(rd(x1))))))
P(0(x1)) → 0¹(s(s(p(x1))))
RD(0(x1)) → 0¹(0(rd(x1)))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
G(x1) → I(s(half(x1)))
I(x1) → P(x1)
RD(0(x1)) → 0¹(rd(x1))
RD(0(x1)) → 0¹(0(0(0(0(0(rd(x1)))))))
I(x1) → F(p(x1))
RD(0(x1)) → 0¹(0(0(0(rd(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F(s(x1)) → G(s(x1))
G(x1) → HALF(x1)
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(s(s(x1))) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
HALF(0(x1)) → 0¹(s(s(half(x1))))
RD(0(x1)) → RD(x1)
RD(0(x1)) → 0¹(0(0(rd(x1))))
P(0(x1)) → P(x1)
HALF(0(x1)) → HALF(x1)
RD(0(x1)) → 0¹(0(0(0(0(rd(x1))))))
P(0(x1)) → 0¹(s(s(p(x1))))
RD(0(x1)) → 0¹(0(rd(x1)))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
G(x1) → I(s(half(x1)))
I(x1) → P(x1)
RD(0(x1)) → 0¹(rd(x1))
RD(0(x1)) → 0¹(0(0(0(0(0(rd(x1)))))))
I(x1) → F(p(x1))
RD(0(x1)) → 0¹(0(0(0(rd(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 12 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

RD(0(x1)) → RD(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

RD(0(x1)) → RD(x1)

R is empty.
Q is empty.
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.

The following dependency pairs can be deleted:

RD(0(x1)) → RD(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 2·x₁
POL(RD(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

RD(0(x1)) → RD(x1)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 2·x₁
POL(P(x₁)) = 2·x₁
POL(p(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

HALF(0(x1)) → HALF(x1)
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

HALF(0(x1)) → HALF(x1)
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)
0(x1) → x1

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

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

HALF(0(x1)) → HALF(x1)
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)
0(x1) → x1

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:

HALF(0(x1)) → HALF(x1)

Strictly oriented rules of the TRS R:

0(x1) → x1

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 2 + 2·x₁
POL(HALF(x₁)) = 2·x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

The set Q consists of the following terms:

p(s(x0))
p(0(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule HALF(s(s(x1))) → HALF(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:

HALF(s(s(x1))) → HALF(p(s(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(s(x1)))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(s(x1)))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule HALF(s(s(x1))) → HALF(p(s(x1))) at position [0] we obtained the following new rules:

HALF(s(s(x1))) → HALF(x1)

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(x1)

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(x1)

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

p(s(x0))
p(0(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.

p(s(x0))
p(0(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(HALF(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ UsableRulesReductionPairsProof

                                                  ↳ QDP

                                                    ↳ PisEmptyProof

          ↳ QDP

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → g(s(x1))
g(x1) → i(s(half(x1)))
i(x1) → f(p(x1))
half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
0(x1) → x1
rd(0(x1)) → 0(0(0(0(0(0(rd(x1)))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

            ↳ UsableRulesProof

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)
0(x1) → x1

0(x1) → x1

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 1 + 2·x₁
POL(F(x₁)) = x₁
POL(G(x₁)) = x₁
POL(I(x₁)) = x₁
POL(half(x₁)) = x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

            ↳ UsableRulesProof

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

            ↳ UsableRulesProof

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

The set Q consists of the following terms:

half(0(x0))
half(s(s(x0)))
p(s(x0))
p(0(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QDPOrderProof

            ↳ UsableRulesProof

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))

The set Q consists of the following terms:

half(0(x0))
half(s(s(x0)))
p(s(x0))
p(0(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(s(x1)) → G(s(x1))

The remaining pairs can at least be oriented weakly.

I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(half(x₁)) = (1/2)x_1
POL(s(x₁)) = 1 + (4)x_1
POL(G(x₁)) = 1/4 + (1/2)x_1
POL(p(x₁)) = (1/4)x_1
POL(0(x₁)) = 0
POL(I(x₁)) = (1/4)x_1
POL(F(x₁)) = x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))

The set Q consists of the following terms:

half(0(x0))
half(s(s(x0)))
p(s(x0))
p(0(x0))

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.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

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

F(s(x1)) → G(s(x1))
I(x1) → F(p(x1))
G(x1) → I(s(half(x1)))

The TRS R consists of the following rules:

half(0(x1)) → 0(s(s(half(x1))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)
0(x1) → x1

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