Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

Q is empty.

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

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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

0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)

The set Q is empty.

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

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

0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)

Q is empty.

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

F(s(x1)) → G(q(i(x1)))
G(x1) → P(x1)
P(0(x1)) → P(x1)
F(s(x1)) → Q(i(x1))
Q(i(x1)) → Q(s(x1))
G(x1) → F(p(p(x1)))
F(s(x1)) → I(x1)
P(s(s(x1))) → P(s(x1))
G(x1) → P(p(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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

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

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

F(s(x1)) → G(q(i(x1)))
G(x1) → P(x1)
P(0(x1)) → P(x1)
F(s(x1)) → Q(i(x1))
Q(i(x1)) → Q(s(x1))
G(x1) → F(p(p(x1)))
F(s(x1)) → I(x1)
P(s(s(x1))) → P(s(x1))
G(x1) → P(p(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → 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 3 SCCs with 5 less nodes.

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

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

P(s(s(x1))) → P(s(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
            ↳ MNOCProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

P(s(s(x1))) → P(s(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
            ↳ MNOCProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

P(s(s(x1))) → P(s(x1))

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

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(x0)



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

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

P(s(s(x1))) → P(s(x1))

R is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

P(s(s(x1))) → P(s(x1))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

P(s(s(x1))) → P(s(x1))

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

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(x0)



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

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

P(s(s(x1))) → P(s(x1))

R is empty.
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:

P(s(s(x1))) → P(s(x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1)) = 2·x1   
POL(s(x1)) = 1 + 2·x1   



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

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

P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
            ↳ MNOCProof
          ↳ QDP
  ↳ QTRS Reverse

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

P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
            ↳ MNOCProof
          ↳ QDP
  ↳ QTRS Reverse

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

P(0(x1)) → P(x1)

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

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(x0)



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

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

P(0(x1)) → P(x1)

R is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
          ↳ QDP
  ↳ QTRS Reverse

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

P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
          ↳ QDP
  ↳ QTRS Reverse

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

P(0(x1)) → P(x1)

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

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(x0)



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

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

P(0(x1)) → P(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

P(0(x1)) → P(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x1)) = 2·x1   
POL(P(x1)) = 2·x1   



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

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
            ↳ MNOCProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
            ↳ MNOCProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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.

f(s(x0))
g(x0)



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

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(s(x0))
i(x0)

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

F(s(x1)) → G(q(s(x1)))



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

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

F(s(x1)) → G(q(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
            ↳ MNOCProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(q(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(s(x0))
i(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.

i(x0)



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

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

F(s(x1)) → G(q(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(s(x0))

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

F(s(x1)) → G(s(s(x1)))



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

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

F(s(x1)) → G(s(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
            ↳ MNOCProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(s(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(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.

q(s(x0))



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

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

F(s(x1)) → G(s(s(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))

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

G(0(x0)) → F(p(s(s(0(s(s(p(x0))))))))
G(s(s(x0))) → F(p(s(p(s(x0)))))
G(s(0(x0))) → F(p(0(x0)))



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

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

F(s(x1)) → G(s(s(x1)))
G(s(s(x0))) → F(p(s(p(s(x0)))))
G(0(x0)) → F(p(s(s(0(s(s(p(x0))))))))
G(s(0(x0))) → F(p(0(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

F(s(x1)) → G(s(s(x1)))
G(s(s(x0))) → F(p(s(p(s(x0)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
                                              ↳ QDP
                                                ↳ Narrowing
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ UsableRulesProof
                                                          ↳ QDP
                                                            ↳ QDPOrderProof
            ↳ MNOCProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(s(s(x1)))
G(s(s(x0))) → F(p(s(p(s(x0)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

G(s(s(x0))) → F(p(s(p(s(x0)))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( p(x1) ) = max{0, x1 - 1}


POL( s(x1) ) = x1 + 1


POL( G(x1) ) = max{0, x1 - 1}


POL( 0(x1) ) = max{0, -1}


POL( F(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

G(s(s(x0))) → F(p(s(p(s(x0)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(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 use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

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

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
  ↳ QTRS Reverse

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
f(s(x0))
g(x0)
q(s(x0))
i(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.

f(s(x0))
g(x0)



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

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(0(x0))
p(s(0(x0)))
p(s(s(x0)))
q(s(x0))
i(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ MNOCProof
            ↳ MNOCProof
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ MNOCProof
                          ↳ QDP
  ↳ QTRS Reverse

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

F(s(x1)) → G(q(i(x1)))
G(x1) → F(p(p(x1)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

p(0(x1)) → s(s(0(s(s(p(x1))))))
p(s(0(x1))) → 0(x1)
p(s(s(x1))) → s(p(s(x1)))
f(s(x1)) → g(q(i(x1)))
g(x1) → f(p(p(x1)))
q(i(x1)) → q(s(x1))
q(s(x1)) → s(s(x1))
i(x1) → s(x1)

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

0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)

The set Q is empty.

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

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

0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)

Q is empty.