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:

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

Q is empty.

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

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

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

0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))

Q is empty.

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

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

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

0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS
  ↳ RRRPoloQTRSProof

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

0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))

Q is empty.

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

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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:

q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
Used ordering:
Polynomial interpretation [25]:

POL(0(x1)) = 1 + 2·x1   
POL(p(x1)) = x1   
POL(q(x1)) = 2·x1   
POL(r(x1)) = 2·x1   
POL(s(x1)) = x1   




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

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

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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:

Q(s(x1)) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
R(s(x1)) → P(s(p(s(s(q(p(s(p(s(x1))))))))))
Q(s(x1)) → P(p(s(s(x1))))
R(s(x1)) → P(s(x1))
R(s(x1)) → P(s(p(s(x1))))
Q(s(x1)) → P(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
R(s(x1)) → Q(p(s(p(s(x1)))))
R(s(x1)) → P(s(s(q(p(s(p(s(x1))))))))
Q(s(x1)) → R(p(p(s(s(x1)))))
Q(s(x1)) → P(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

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

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

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

Q(s(x1)) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
R(s(x1)) → P(s(p(s(s(q(p(s(p(s(x1))))))))))
Q(s(x1)) → P(p(s(s(x1))))
R(s(x1)) → P(s(x1))
R(s(x1)) → P(s(p(s(x1))))
Q(s(x1)) → P(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
R(s(x1)) → Q(p(s(p(s(x1)))))
R(s(x1)) → P(s(s(q(p(s(p(s(x1))))))))
Q(s(x1)) → R(p(p(s(s(x1)))))
Q(s(x1)) → P(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

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 8 less nodes.

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

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

P(p(s(x1))) → P(x1)

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                ↳ UsableRulesProof
              ↳ QDP

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

P(p(s(x1))) → P(x1)

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

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

P(p(s(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)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
QDP
                    ↳ UsableRulesProof
                ↳ MNOCProof

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

Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

The set Q consists of the following terms:

q(s(x0))
r(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
QDP
                        ↳ QReductionProof
                ↳ MNOCProof

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

Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

q(s(x0))
r(s(x0))
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.

q(s(x0))
r(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
QDP
                            ↳ Rewriting
                ↳ MNOCProof

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

R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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 Q(s(x1)) → R(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:

Q(s(x1)) → R(p(s(x1)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
QDP
                                ↳ UsableRulesProof
                ↳ MNOCProof

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

Q(s(x1)) → R(p(s(x1)))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ Rewriting
                ↳ MNOCProof

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

Q(s(x1)) → R(p(s(x1)))
R(s(x1)) → Q(p(s(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 R(s(x1)) → Q(p(s(p(s(x1))))) at position [0] we obtained the following new rules:

R(s(x1)) → Q(p(s(x1)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
QDP
                                        ↳ Rewriting
                ↳ MNOCProof

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

R(s(x1)) → Q(p(s(x1)))
Q(s(x1)) → R(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 Q(s(x1)) → R(p(s(x1))) at position [0] we obtained the following new rules:

Q(s(x1)) → R(x1)



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
QDP
                                            ↳ Rewriting
                ↳ MNOCProof

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

R(s(x1)) → Q(p(s(x1)))
Q(s(x1)) → R(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 R(s(x1)) → Q(p(s(x1))) at position [0] we obtained the following new rules:

R(s(x1)) → Q(x1)



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
QDP
                                                ↳ UsableRulesProof
                ↳ MNOCProof

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

Q(s(x1)) → R(x1)
R(s(x1)) → Q(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ UsableRulesProof
QDP
                                                    ↳ QReductionProof
                ↳ MNOCProof

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

Q(s(x1)) → R(x1)
R(s(x1)) → Q(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ UsableRulesProof
                                                  ↳ QDP
                                                    ↳ QReductionProof
QDP
                                                        ↳ UsableRulesReductionPairsProof
                ↳ MNOCProof

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

Q(s(x1)) → R(x1)
R(s(x1)) → Q(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:

Q(s(x1)) → R(x1)
R(s(x1)) → Q(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(Q(x1)) = 2 + x1   
POL(R(x1)) = 1 + 2·x1   
POL(s(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ UsableRulesProof
                                                  ↳ QDP
                                                    ↳ QReductionProof
                                                      ↳ QDP
                                                        ↳ UsableRulesReductionPairsProof
QDP
                                                            ↳ PisEmptyProof
                ↳ MNOCProof

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 use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

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

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

Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))

The set Q consists of the following terms:

q(s(x0))
r(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ MNOCProof
                ↳ MNOCProof
                  ↳ QDP
                    ↳ UsableRulesProof
QDP
                        ↳ QReductionProof

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

Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

q(s(x0))
r(s(x0))
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.

q(s(x0))
r(s(x0))



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

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

R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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.
We use the modular non-overlap check [17] to decrease Q to the empty set.

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

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

Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))

The TRS R consists of the following rules:

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

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