Termination w.r.t. Q proof of /tmp/tmpxpLjKE/z074.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:

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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(n(x1)) → s(r(x1))
t(n(u(x1))) → t(c(r(x1)))

Used ordering:
Polynomial interpretation [25]:

POL(b(x₁)) = x₁
POL(c(x₁)) = x₁
POL(n(x₁)) = 1 + 2·x₁
POL(r(x₁)) = x₁
POL(s(x₁)) = x₁
POL(t(x₁)) = x₁
POL(u(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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)) → s(r(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))

Used ordering:
Polynomial interpretation [25]:

POL(b(x₁)) = 2·x₁
POL(c(x₁)) = x₁
POL(n(x₁)) = 2·x₁
POL(r(x₁)) = 2 + x₁
POL(s(x₁)) = x₁
POL(t(x₁)) = x₁
POL(u(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:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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)
C(u(x1)) → C(x1)
R(s(x1)) → R(x1)
R(u(x1)) → R(x1)
R(s(x1)) → S(r(x1))
R(b(x1)) → S(b(x1))
T(s(u(x1))) → T(c(r(x1)))
C(s(x1)) → S(c(x1))
C(r(x1)) → R(c(x1))
T(s(u(x1))) → C(r(x1))
C(s(x1)) → C(x1)
S(u(x1)) → S(x1)
T(s(u(x1))) → R(x1)
C(r(x1)) → C(x1)

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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)
C(u(x1)) → C(x1)
R(s(x1)) → R(x1)
R(u(x1)) → R(x1)
R(s(x1)) → S(r(x1))
R(b(x1)) → S(b(x1))
T(s(u(x1))) → T(c(r(x1)))
C(s(x1)) → S(c(x1))
C(r(x1)) → R(c(x1))
T(s(u(x1))) → C(r(x1))
C(s(x1)) → C(x1)
S(u(x1)) → S(x1)
T(s(u(x1))) → R(x1)
C(r(x1)) → C(x1)

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

S(u(x1)) → S(x1)

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(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

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

S(u(x1)) → S(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

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

S(u(x1)) → S(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:

S(u(x1)) → S(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(S(x₁)) = 2·x₁
POL(u(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

                  ↳ 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

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

R(u(x1)) → R(x1)
R(s(x1)) → R(x1)

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                    ↳ UsableRulesProof

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

R(u(x1)) → R(x1)
R(s(x1)) → R(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(R(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁
POL(u(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

                    ↳ UsableRulesProof

                  ↳ 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.
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

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

R(u(x1)) → R(x1)
R(s(x1)) → R(x1)

R is empty.
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

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ 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(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ 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(u(x1)) → C(x1)
C(s(x1)) → C(x1)
C(r(x1)) → C(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x₁)) = 2·x₁
POL(n(x₁)) = 2·x₁
POL(r(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁
POL(u(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ 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

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
t(s(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)
s(u(x1)) → u(s(x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x1))) → T(c(r(x1)))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)
s(u(x1)) → u(s(x1))

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

T(s(u(b(x0)))) → T(c(u(s(b(x0)))))
T(s(u(u(x0)))) → T(c(u(r(x0))))
T(s(u(s(x0)))) → T(c(s(r(x0))))
T(s(u(x0))) → T(r(c(x0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x0))) → T(r(c(x0)))
T(s(u(b(x0)))) → T(c(u(s(b(x0)))))
T(s(u(s(x0)))) → T(c(s(r(x0))))
T(s(u(u(x0)))) → T(c(u(r(x0))))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)
s(u(x1)) → u(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 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x0))) → T(r(c(x0)))
T(s(u(s(x0)))) → T(c(s(r(x0))))
T(s(u(u(x0)))) → T(c(u(r(x0))))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)
s(u(x1)) → u(s(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(s(u(u(x0)))) → T(c(u(r(x0))))

The remaining pairs can at least be oriented weakly.

T(s(u(x0))) → T(r(c(x0)))
T(s(u(s(x0)))) → T(c(s(r(x0))))

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

POL( n(x₁) ) = max{0, -1}

POL( c(x₁) ) = x₁

POL( T(x₁) ) = x₁

POL( u(x₁) ) = max{0, -1}

POL( s(x₁) ) = 1

POL( r(x₁) ) = 1

POL( b(x₁) ) = max{0, x₁ - 1}

The following usable rules [17] were oriented:

r(b(x1)) → u(s(b(x1)))
r(s(x1)) → s(r(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
s(u(x1)) → u(s(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T(s(u(x0))) → T(r(c(x0)))
T(s(u(s(x0)))) → T(c(s(r(x0))))

The TRS R consists of the following rules:

r(s(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)
s(u(x1)) → u(s(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.n: 1
c: 1
T: 0
u: x₀
s: x₀
r: x₀
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.0(s.0(u.0(x0))) → T.1(r.1(c.0(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))
T.0(s.0(u.0(s.0(x0)))) → T.1(c.0(s.0(r.0(x0))))

The TRS R consists of the following rules:

r.0(u.0(x1)) → u.0(r.0(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
r.1(s.1(x1)) → s.1(r.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
r.0(s.0(x1)) → s.0(r.0(x1))
c.0(r.0(x1)) → r.1(c.0(x1))
c.0(u.0(x1)) → u.1(c.0(x1))
r.0(b.0(x1)) → u.0(s.0(b.0(x1)))
r.0(b.1(x1)) → u.0(s.0(b.1(x1)))
c.1(n.0(x1)) → n.0(x1)
c.0(s.0(x1)) → s.1(c.0(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(n.0(x1)) → n.1(c.0(x1))
c.1(s.1(x1)) → s.1(c.1(x1))
s.0(u.0(x1)) → u.0(s.0(x1))
c.1(n.1(x1)) → n.1(x1)
s.1(u.1(x1)) → u.1(s.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

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.0(s.0(u.0(x0))) → T.1(r.1(c.0(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))
T.0(s.0(u.0(s.0(x0)))) → T.1(c.0(s.0(r.0(x0))))

The TRS R consists of the following rules:

r.0(u.0(x1)) → u.0(r.0(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
r.1(s.1(x1)) → s.1(r.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
r.0(s.0(x1)) → s.0(r.0(x1))
c.0(r.0(x1)) → r.1(c.0(x1))
c.0(u.0(x1)) → u.1(c.0(x1))
r.0(b.0(x1)) → u.0(s.0(b.0(x1)))
r.0(b.1(x1)) → u.0(s.0(b.1(x1)))
c.1(n.0(x1)) → n.0(x1)
c.0(s.0(x1)) → s.1(c.0(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(n.0(x1)) → n.1(c.0(x1))
c.1(s.1(x1)) → s.1(c.1(x1))
s.0(u.0(x1)) → u.0(s.0(x1))
c.1(n.1(x1)) → n.1(x1)
s.1(u.1(x1)) → u.1(s.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

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))

The TRS R consists of the following rules:

r.0(u.0(x1)) → u.0(r.0(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
r.1(s.1(x1)) → s.1(r.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
r.0(s.0(x1)) → s.0(r.0(x1))
c.0(r.0(x1)) → r.1(c.0(x1))
c.0(u.0(x1)) → u.1(c.0(x1))
r.0(b.0(x1)) → u.0(s.0(b.0(x1)))
r.0(b.1(x1)) → u.0(s.0(b.1(x1)))
c.1(n.0(x1)) → n.0(x1)
c.0(s.0(x1)) → s.1(c.0(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(n.0(x1)) → n.1(c.0(x1))
c.1(s.1(x1)) → s.1(c.1(x1))
s.0(u.0(x1)) → u.0(s.0(x1))
c.1(n.1(x1)) → n.1(x1)
s.1(u.1(x1)) → u.1(s.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(n.0(x1)) → n.1(c.0(x1))
c.0(r.0(x1)) → r.1(c.0(x1))
c.0(u.0(x1)) → u.1(c.0(x1))
c.0(s.0(x1)) → s.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(n.0(x₁)) = 1 + x₁
POL(n.1(x₁)) = x₁
POL(r.0(x₁)) = 1 + x₁
POL(r.1(x₁)) = x₁
POL(s.0(x₁)) = 1 + x₁
POL(s.1(x₁)) = x₁
POL(u.0(x₁)) = 1 + x₁
POL(u.1(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

                                              ↳ QDP

                                                ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))

The TRS R consists of the following rules:

r.1(s.1(x1)) → s.1(r.1(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
s.1(u.1(x1)) → u.1(s.1(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
c.1(n.0(x1)) → n.0(x1)
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(s.1(x1)) → s.1(c.1(x1))
c.1(n.1(x1)) → n.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(n.0(x1)) → n.0(x1)
c.1(n.1(x1)) → n.1(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x₁)) = x₁
POL(c.1(x₁)) = 1 + x₁
POL(n.0(x₁)) = x₁
POL(n.1(x₁)) = x₁
POL(r.1(x₁)) = x₁
POL(s.1(x₁)) = x₁
POL(u.1(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

                                              ↳ QDP

                                                ↳ RuleRemovalProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))

The TRS R consists of the following rules:

r.1(s.1(x1)) → s.1(r.1(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
s.1(u.1(x1)) → u.1(s.1(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(s.1(x1)) → s.1(c.1(x1))

T.1(s.1(u.1(x0))) → T.1(r.1(c.1(x0)))
T.1(s.1(u.1(s.1(x0)))) → T.1(c.1(s.1(r.1(x0))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x₁)) = x₁
POL(c.1(x₁)) = x₁
POL(n.1(x₁)) = x₁
POL(r.1(x₁)) = x₁
POL(s.1(x₁)) = x₁
POL(u.1(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ SemLabProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

                                              ↳ QDP

                                                ↳ RuleRemovalProof

                                                  ↳ QDP

                                                    ↳ RuleRemovalProof

                                                      ↳ QDP

                                                        ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

r.1(s.1(x1)) → s.1(r.1(x1))
r.1(u.1(x1)) → u.1(r.1(x1))
s.1(u.1(x1)) → u.1(s.1(x1))
c.1(u.1(x1)) → u.1(c.1(x1))
c.1(r.1(x1)) → r.1(c.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(s.1(x1)) → s.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 have reversed the following QTRS:
The set of rules R is

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

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

r(r(x)) → r(s(x))
s(r(x)) → r(s(x))
n(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(r(t(x))) → r(c(t(x)))
u(s(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:

r(r(x)) → r(s(x))
s(r(x)) → r(s(x))
n(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(r(t(x))) → r(c(t(x)))
u(s(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)

Q is empty.

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

r(r(x1)) → s(r(x1))
r(s(x1)) → s(r(x1))
r(n(x1)) → s(r(x1))
r(b(x1)) → u(s(b(x1)))
r(u(x1)) → u(r(x1))
s(u(x1)) → u(s(x1))
n(u(x1)) → u(n(x1))
t(r(u(x1))) → t(c(r(x1)))
t(s(u(x1))) → t(c(r(x1)))
t(n(u(x1))) → t(c(r(x1)))
c(u(x1)) → u(c(x1))
c(s(x1)) → s(c(x1))
c(r(x1)) → r(c(x1))
c(n(x1)) → n(c(x1))
c(n(x1)) → n(x1)

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

r(r(x)) → r(s(x))
s(r(x)) → r(s(x))
n(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(r(t(x))) → r(c(t(x)))
u(s(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:

r(r(x)) → r(s(x))
s(r(x)) → r(s(x))
n(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(r(t(x))) → r(c(t(x)))
u(s(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)

Q is empty.