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:

R(x1) → r(x1)
r(p(x1)) → p(p(r(P(x1))))
r(r(x1)) → x1
r(P(P(x1))) → P(P(r(x1)))
p(P(x1)) → x1
P(p(x1)) → x1
r(R(x1)) → x1
R(r(x1)) → x1

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

R(x1) → r(x1)
r(p(x1)) → p(p(r(P(x1))))
r(r(x1)) → x1
r(P(P(x1))) → P(P(r(x1)))
p(P(x1)) → x1
P(p(x1)) → x1
r(R(x1)) → x1
R(r(x1)) → x1

Q is empty.

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

R(x1) → r(x1)
r(p(x1)) → p(p(r(P(x1))))
r(r(x1)) → x1
r(P(P(x1))) → P(P(r(x1)))
p(P(x1)) → x1
P(p(x1)) → x1
r(R(x1)) → x1
R(r(x1)) → 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:

r(r(x1)) → x1
p(P(x1)) → x1
P(p(x1)) → x1
r(R(x1)) → x1
R(r(x1)) → x1
Used ordering:
Polynomial interpretation [25]:

POL(P(x1)) = x1   
POL(R(x1)) = 2 + 2·x1   
POL(p(x1)) = 1 + x1   
POL(r(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

R(x1) → r(x1)
r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(x1)))

Q is empty.

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

R(x1) → r(x1)
r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(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:

R(x1) → r(x1)
Used ordering:
Polynomial interpretation [25]:

POL(P(x1)) = x1   
POL(R(x1)) = 1 + 2·x1   
POL(p(x1)) = x1   
POL(r(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence

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

r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(x1)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof

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

r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(x1)))

The set Q consists of the following terms:

r(p(x0))
r(P(P(x0)))


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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))

The TRS R consists of the following rules:

r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(x1)))

The set Q consists of the following terms:

r(p(x0))
r(P(P(x0)))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                  ↳ UsableRulesProof

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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))

The TRS R consists of the following rules:

r(p(x1)) → p(p(r(P(x1))))
r(P(P(x1))) → P(P(r(x1)))

The set Q consists of the following terms:

r(p(x0))
r(P(P(x0)))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                  ↳ UsableRulesProof

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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))

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

r(p(x0))
r(P(P(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.

r(p(x0))
r(P(P(x0)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                        ↳ QDP
                  ↳ UsableRulesProof

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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof

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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))

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

r(p(x0))
r(P(P(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.

r(p(x0))
r(P(P(x0)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                        ↳ QDP
                          ↳ UsableRulesReductionPairsProof

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

R(P(P(x1))) → R(x1)
R(p(x1)) → R(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:

R(P(P(x1))) → R(x1)
R(p(x1)) → R(P(x1))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1)) = 1 + x1   
POL(R(x1)) = 2·x1   
POL(p(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ Overlay + Local Confluence
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                        ↳ QDP
                          ↳ UsableRulesReductionPairsProof
                            ↳ QDP
                              ↳ PisEmptyProof

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.