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:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

Q is empty.


QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

Q is empty.

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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

b(b(b(b(x1)))) → a(x1)
Used ordering:
Polynomial interpretation [25]:

POL(a(x1)) = 2 + x1   
POL(b(x1)) = 1 + x1   
POL(f(x1)) = x1   
POL(g(x1)) = 2·x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
QTRS
      ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))

Q is empty.

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))

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

a(a(a(x1))) → b(a(a(b(x1))))
Used ordering:
Polynomial interpretation [25]:

POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
QTRS
          ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))

Q is empty.

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))

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

b(a(x1)) → a(b(x1))
Used ordering:
Polynomial interpretation [25]:

POL(a(x1)) = 1 + x1   
POL(b(x1)) = 2 + 2·x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
QTRS
              ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))

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(h(x1)) → H(f(s(h(x1))))
F(h(x1)) → F(s(h(x1)))
F(f(s(s(x1)))) → F(f(x1))
F(s(s(s(x1)))) → H(x1)
F(s(s(s(x1)))) → F(s(h(x1)))
G(h(x1)) → F(s(x1))
G(h(x1)) → G(f(s(x1)))
F(s(s(s(x1)))) → H(f(s(h(x1))))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
QDP
                  ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

F(h(x1)) → H(f(s(h(x1))))
F(h(x1)) → F(s(h(x1)))
F(f(s(s(x1)))) → F(f(x1))
F(s(s(s(x1)))) → H(x1)
F(s(s(s(x1)))) → F(s(h(x1)))
G(h(x1)) → F(s(x1))
G(h(x1)) → G(f(s(x1)))
F(s(s(s(x1)))) → H(f(s(h(x1))))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ MNOCProof
                        ↳ UsableRulesProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

h(x1) → 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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
QDP
                                ↳ MNOCProof
                        ↳ UsableRulesProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

h(x1) → x1

The set Q consists of the following terms:

h(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
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
                              ↳ QDP
                                ↳ MNOCProof
QDP
                        ↳ UsableRulesProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

h(x1) → x1

Q is empty.
We have to consider all (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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
QDP
                            ↳ MNOCProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

h(x1) → 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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
QDP
                                ↳ RuleRemovalProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

F(s(s(s(x1)))) → F(s(h(x1)))

The TRS R consists of the following rules:

h(x1) → x1

The set Q consists of the following terms:

h(x0)

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:

F(s(s(s(x1)))) → F(s(h(x1)))

Strictly oriented rules of the TRS R:

h(x1) → x1

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = 2·x1   
POL(h(x1)) = 1 + x1   
POL(s(x1)) = 1 + x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
                              ↳ QDP
                                ↳ RuleRemovalProof
QDP
                                    ↳ PisEmptyProof
                      ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

h(x0)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                      ↳ QDP
  ↳ QTRS Reverse

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

F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                        ↳ UsableRulesProof
                      ↳ QDP
  ↳ QTRS Reverse

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

F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
QDP
                            ↳ RuleRemovalProof
                      ↳ QDP
  ↳ QTRS Reverse

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

F(f(s(s(x1)))) → F(f(x1))
F(f(s(s(x1)))) → F(x1)

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(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:

F(f(s(s(x1)))) → F(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = x1   
POL(f(x1)) = 2 + 2·x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
QDP
                                ↳ QDPToSRSProof
                      ↳ QDP
  ↳ QTRS Reverse

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

F(f(s(s(x1)))) → F(f(x1))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
QTRS
                                    ↳ QTRS Reverse
                      ↳ QDP
  ↳ QTRS Reverse

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

f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
F(f(s(s(x1)))) → F(f(x1))

Q is empty.

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

f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
h(x1) → x1
F(f(s(s(x1)))) → F(f(x1))

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

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                      ↳ QDP
  ↳ QTRS Reverse

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

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

Q is empty.

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

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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

f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
QTRS
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                      ↳ QDP
  ↳ QTRS Reverse

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

f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))

Q is empty.

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

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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

f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(x))

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
QTRS
                                        ↳ DependencyPairsProof
                      ↳ QDP
  ↳ QTRS Reverse

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

f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
f(f(s(s(x)))) → s(s(s(f(f(x)))))
h(x) → x
F(f(s(s(x)))) → F(f(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:

H(f(x)) → H(x)
H(f(x)) → H(s(f(h(x))))
S(s(s(f(x)))) → S(f(h(x)))
H(f(x)) → S(f(h(x)))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
S(s(s(f(x)))) → H(x)
S(s(s(f(x)))) → H(s(f(h(x))))

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
QDP
                                            ↳ DependencyGraphProof
                      ↳ QDP
  ↳ QTRS Reverse

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

H(f(x)) → H(x)
H(f(x)) → H(s(f(h(x))))
S(s(s(f(x)))) → S(f(h(x)))
H(f(x)) → S(f(h(x)))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))
S(s(s(f(x)))) → H(x)
S(s(s(f(x)))) → H(s(f(h(x))))

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
QDP
                                                  ↳ UsableRulesProof
                                                  ↳ UsableRulesProof
                                                ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

H(f(x)) → H(x)

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ UsableRulesProof
QDP
                                                  ↳ UsableRulesProof
                                                ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

H(f(x)) → H(x)

R is empty.
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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                  ↳ UsableRulesProof
QDP
                                                      ↳ UsableRulesReductionPairsProof
                                                ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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

H(f(x)) → H(x)

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:

H(f(x)) → H(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(H(x1)) = 2·x1   
POL(f(x1)) = 2·x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ UsableRulesReductionPairsProof
QDP
                                                          ↳ PisEmptyProof
                                                ↳ QDP
                      ↳ QDP
  ↳ QTRS Reverse

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
QDP
                                                  ↳ RuleRemovalProof
                      ↳ QDP
  ↳ QTRS Reverse

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

S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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:

S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)
S(s(f(f(x)))) → S(s(x))


Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = x1   
POL(S(x1)) = 2·x1   
POL(f(x1)) = 2 + x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ QDPToSRSProof
                                  ↳ QTRS
                                    ↳ QTRS Reverse
                                      ↳ QTRS
                                        ↳ QTRS Reverse
                                        ↳ QTRS Reverse
                                        ↳ DependencyPairsProof
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
QDP
                                                      ↳ PisEmptyProof
                      ↳ QDP
  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(x) → x
s(s(f(F(x)))) → f(F(x))

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
  ↳ QTRS Reverse

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                        ↳ UsableRulesProof
  ↳ QTRS Reverse

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))
h(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
QDP
                            ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))
h(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:

h(x1) → x1

Used ordering: POLO with Polynomial interpretation [25]:

POL(G(x1)) = x1   
POL(f(x1)) = x1   
POL(h(x1)) = 2 + 2·x1   
POL(s(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
QDP
                                ↳ MNOCProof
  ↳ QTRS Reverse

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ MNOCProof
QDP
                                    ↳ Narrowing
  ↳ QTRS Reverse

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

G(h(x1)) → G(f(s(x1)))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))

The set Q consists of the following terms:

f(s(s(s(x0))))

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

G(h(s(s(x0)))) → G(h(f(s(h(x0)))))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ MNOCProof
                                  ↳ QDP
                                    ↳ Narrowing
QDP
                                        ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

G(h(s(s(x0)))) → G(h(f(s(h(x0)))))

The TRS R consists of the following rules:

f(s(s(s(x1)))) → h(f(s(h(x1))))

The set Q consists of the following terms:

f(s(s(s(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 have reversed the following QTRS:
The set of rules R is

g(h(x1)) → g(f(s(x1)))
f(s(s(s(x1)))) → h(f(s(h(x1))))
f(h(x1)) → h(f(s(h(x1))))
h(x1) → x1
f(f(s(s(x1)))) → s(s(s(f(f(x1)))))
b(a(x1)) → a(b(x1))
a(a(a(x1))) → b(a(a(b(x1))))
b(b(b(b(x1)))) → a(x1)

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

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
QTRS

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

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)

Q is empty.