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)))))))

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
          ↳ 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)

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:

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

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
          ↳ 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)))))))

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
          ↳ 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

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
          ↳ 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))

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
                                ↳ 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))

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
                                ↳ 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)

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:

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)))))))

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
          ↳ 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

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 rules of the TRS R:

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))

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
          ↳ 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))
G(x1) → I(s(half(x1)))
The remaining pairs can at least be oriented weakly.

I(x1) → F(p(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(half(x1)) = (1/2)x_1   
POL(s(x1)) = 4 + (4)x_1   
POL(G(x1)) = 4 + x_1   
POL(p(x1)) = (1/4)x_1   
POL(0(x1)) = 0   
POL(I(x1)) = 1 + (1/2)x_1   
POL(F(x1)) = 1 + (2)x_1   
The value of delta used in the strict ordering is 1.
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))

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 1 less node.
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.