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:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

Q is empty.

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

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

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

s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → 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:

s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → x

Q is empty.

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

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

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

s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → 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:

s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → 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:

P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → 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:

P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → 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 11 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ 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:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(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
            ↳ UsableRulesProof
QDP
                ↳ UsableRulesReductionPairsProof
            ↳ 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.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
QDP
                    ↳ PisEmptyProof
            ↳ UsableRulesProof
          ↳ 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

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.

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

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
QDP
                ↳ UsableRulesProof
            ↳ MNOCProof

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

The set Q consists of the following terms:

a(s(x0))
b(s(x0))
c(s(x0))
p(s(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
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
            ↳ MNOCProof

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

a(s(x0))
b(s(x0))
c(s(x0))
p(s(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.

a(s(x0))
b(s(x0))
c(s(x0))



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

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(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))

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ MNOCProof
QDP
            ↳ MNOCProof

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

Q is empty.
We have to consider all (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
            ↳ MNOCProof
            ↳ MNOCProof
QDP
                ↳ UsableRulesProof

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1

The set Q consists of the following terms:

a(s(x0))
b(s(x0))
c(s(x0))
p(s(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
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

a(s(x0))
b(s(x0))
c(s(x0))
p(s(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.

a(s(x0))
b(s(x0))
c(s(x0))



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

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

B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(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))

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

B(s(x1)) → C(p(s(x1)))



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

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

C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
B(s(x1)) → C(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))

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

C(s(x1)) → A(p(s(x1)))



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

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

A(s(x1)) → B(p(p(s(s(x1)))))
C(s(x1)) → A(p(s(x1)))
B(s(x1)) → C(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))

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

A(s(x1)) → B(p(s(x1)))



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

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

A(s(x1)) → B(p(s(x1)))
B(s(x1)) → C(p(s(x1)))
C(s(x1)) → A(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))

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

B(s(x1)) → C(x1)



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

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

B(s(x1)) → C(x1)
A(s(x1)) → B(p(s(x1)))
C(s(x1)) → A(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))

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

C(s(x1)) → A(x1)



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

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

B(s(x1)) → C(x1)
C(s(x1)) → A(x1)
A(s(x1)) → B(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))

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

A(s(x1)) → B(x1)



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

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

B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(x1)

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(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
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ UsableRulesProof
QDP
                                                    ↳ QReductionProof

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

B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(x1)

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

p(s(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))



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

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

B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(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:

B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + x1   
POL(B(x1)) = x1   
POL(C(x1)) = 2 + 2·x1   
POL(s(x1)) = 2 + 2·x1   



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