Termination w.r.t. Q proof of /tmp/tmp1QF42P/z075.srs

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:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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:

o(f(x1)) → f(o(x1))

Used ordering:
Polynomial interpretation [25]:

POL(c(x₁)) = x₁
POL(f(x₁)) = 2 + x₁
POL(n(x₁)) = 2 + x₁
POL(o(x₁)) = 1 + 2·x₁
POL(s(x₁)) = 1 + x₁
POL(t(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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:

o(s(x1)) → f(s(x1))

Used ordering:
Polynomial interpretation [25]:

POL(c(x₁)) = x₁
POL(f(x₁)) = 2·x₁
POL(n(x₁)) = 2·x₁
POL(o(x₁)) = 1 + 2·x₁
POL(s(x₁)) = 2·x₁
POL(t(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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:

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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 3 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N(f(x1)) → N(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N(f(x1)) → N(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N(f(x1)) → N(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:

N(f(x1)) → N(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(N(x₁)) = 2·x₁
POL(f(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(f(x1)) → C(x1)
C(o(x1)) → C(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(f(x1)) → C(x1)
C(o(x1)) → C(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x₁)) = 2·x₁
POL(f(x₁)) = 2·x₁
POL(n(x₁)) = 2·x₁
POL(o(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))
T(f(s(x0))) → T(c(f(s(x0))))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))
T(f(s(x0))) → T(c(f(s(x0))))

The TRS R consists of the following rules:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.
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.

T(f(s(x0))) → T(c(f(s(x0))))

The remaining pairs can at least be oriented weakly.

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( s(x₁) ) = x₁ + 1

POL( T(x₁) ) = x₁

POL( n(x₁) ) = x₁

POL( f(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( o(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

n(f(x1)) → f(n(x1))
c(o(x1)) → o(x1)
c(o(x1)) → o(c(x1))
c(n(x1)) → n(c(x1))
c(f(x1)) → f(c(x1))
n(s(x1)) → f(s(x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))

The TRS R consists of the following rules:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 1
f: x₀
n: x₀
T: 0
s: 0
o: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.0(f.0(x0)) → T.1(n.1(c.0(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))
T.0(f.0(f.0(x0))) → T.1(c.0(f.0(n.0(x0))))

The TRS R consists of the following rules:

c.0(n.0(x1)) → n.1(c.0(x1))
n.0(s.1(x1)) → f.0(s.1(x1))
n.1(f.1(x1)) → f.1(n.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(f.1(x1)) → f.1(c.1(x1))
n.0(f.0(x1)) → f.0(n.0(x1))
c.1(o.0(x1)) → o.1(c.0(x1))
n.0(s.0(x1)) → f.0(s.0(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(c.1(x1))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.0(f.0(x0)) → T.1(n.1(c.0(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))
T.0(f.0(f.0(x0))) → T.1(c.0(f.0(n.0(x0))))

The TRS R consists of the following rules:

c.0(n.0(x1)) → n.1(c.0(x1))
n.0(s.1(x1)) → f.0(s.1(x1))
n.1(f.1(x1)) → f.1(n.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(f.1(x1)) → f.1(c.1(x1))
n.0(f.0(x1)) → f.0(n.0(x1))
c.1(o.0(x1)) → o.1(c.0(x1))
n.0(s.0(x1)) → f.0(s.0(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(c.1(x1))

Q is empty.
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

c.0(n.0(x1)) → n.1(c.0(x1))
n.0(s.1(x1)) → f.0(s.1(x1))
n.1(f.1(x1)) → f.1(n.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(f.1(x1)) → f.1(c.1(x1))
n.0(f.0(x1)) → f.0(n.0(x1))
c.1(o.0(x1)) → o.1(c.0(x1))
n.0(s.0(x1)) → f.0(s.0(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(c.1(x1))

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.

No dependency pairs are removed.

The following rules are removed from R:

c.1(o.0(x1)) → o.1(c.0(x1))
c.0(n.0(x1)) → n.1(c.0(x1))
c.0(f.0(x1)) → f.1(c.0(x1))

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x₁)) = x₁
POL(c.0(x₁)) = x₁
POL(c.1(x₁)) = x₁
POL(f.0(x₁)) = 1 + x₁
POL(f.1(x₁)) = x₁
POL(n.0(x₁)) = 1 + x₁
POL(n.1(x₁)) = x₁
POL(o.0(x₁)) = 1 + x₁
POL(o.1(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

n.1(f.1(x1)) → f.1(n.1(x1))
c.1(f.1(x1)) → f.1(c.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(c.1(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

c.1(o.1(x1)) → o.1(x1)
c.1(o.0(x1)) → o.0(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x₁)) = x₁
POL(c.1(x₁)) = 1 + x₁
POL(f.1(x₁)) = 1 + x₁
POL(n.1(x₁)) = x₁
POL(o.0(x₁)) = x₁
POL(o.1(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ RuleRemovalProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

n.1(f.1(x1)) → f.1(n.1(x1))
c.1(f.1(x1)) → f.1(c.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(o.1(x1)) → o.1(c.1(x1))

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x₁)) = x₁
POL(c.1(x₁)) = x₁
POL(f.1(x₁)) = 1 + x₁
POL(n.1(x₁)) = x₁
POL(o.1(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ RuleRemovalProof

                                                  ↳ QDP

                                                    ↳ PisEmptyProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

n.1(f.1(x1)) → f.1(n.1(x1))
c.1(f.1(x1)) → f.1(c.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(o.1(x1)) → o.1(c.1(x1))

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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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

f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
s(o(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
s(o(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)

Q is empty.

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

t(f(x1)) → t(c(n(x1)))
n(f(x1)) → f(n(x1))
o(f(x1)) → f(o(x1))
n(s(x1)) → f(s(x1))
o(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(x1)

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

f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
s(o(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
s(o(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)

Q is empty.