Termination w.r.t. Q proof of /tmp/tmp1NEp87/while1.srs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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

2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))

Q is empty.

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

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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

2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(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:

2¹(c(1(x1))) → R¹(1(x1))
2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
R¹(1(x1)) → 3¹(x1)
R¹(3(x1)) → 3¹(R(x1))
3¹(L(x1)) → 3¹(x1)
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

2¹(c(1(x1))) → R¹(1(x1))
2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
R¹(1(x1)) → 3¹(x1)
R¹(3(x1)) → 3¹(R(x1))
3¹(L(x1)) → 3¹(x1)
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

3¹(L(x1)) → 3¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

3¹(L(x1)) → 3¹(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

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

3¹(L(x1)) → 3¹(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:

3¹(L(x1)) → 3¹(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(3¹(x₁)) = 2·x₁
POL(L(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

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

2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))
R¹(3(x1)) → R¹(x1)
R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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 dependency pairs:

R¹(2(x1)) → 2¹(R(x1))
R¹(2(x1)) → R¹(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 1 + x₁
POL(0¹(x₁)) = 1 + x₁
POL(1(x₁)) = x₁
POL(2(x₁)) = 1 + x₁
POL(2¹(x₁)) = 1 + x₁
POL(3(x₁)) = x₁
POL(L(x₁)) = x₁
POL(R(x₁)) = x₁
POL(R¹(x₁)) = 1 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

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

2¹(L(x1)) → 2¹(x1)
2¹(c(1(x1))) → 0¹(R(1(x1)))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(0(x1))) → 0¹(0(x1))
R¹(3(x1)) → R¹(x1)
0¹(L(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ UsableRulesProof

                    ↳ QDP

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

R¹(3(x1)) → R¹(x1)

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                    ↳ QDP

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

R¹(3(x1)) → R¹(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(3(x₁)) = 2·x₁
POL(R¹(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ PisEmptyProof

                    ↳ QDP

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ Narrowing

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

2¹(L(x1)) → 2¹(x1)
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(x1)) → 2¹(R(x1))
2¹(c(1(x1))) → 0¹(R(1(x1)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 2¹(c(1(x1))) → 0¹(R(1(x1))) at position [0] we obtained the following new rules:

2¹(c(1(x0))) → 0¹(L(3(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

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

2¹(L(x1)) → 2¹(x1)
0¹(L(x1)) → 2¹(R(x1))
2¹(c(0(x1))) → 0¹(0(x1))
2¹(c(1(x0))) → 0¹(L(3(x0)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 0¹(L(x1)) → 2¹(R(x1)) at position [0] we obtained the following new rules:

0¹(L(b(x0))) → 2¹(c(1(b(x0))))
0¹(L(2(x0))) → 2¹(2(R(x0)))
0¹(L(1(x0))) → 2¹(L(3(x0)))
0¹(L(3(x0))) → 2¹(3(R(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

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

2¹(L(x1)) → 2¹(x1)
0¹(L(3(x0))) → 2¹(3(R(x0)))
0¹(L(b(x0))) → 2¹(c(1(b(x0))))
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(2(x0))) → 2¹(2(R(x0)))
2¹(c(1(x0))) → 0¹(L(3(x0)))
0¹(L(1(x0))) → 2¹(L(3(x0)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

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

2¹(L(x1)) → 2¹(x1)
0¹(L(3(x0))) → 2¹(3(R(x0)))
2¹(c(0(x1))) → 0¹(0(x1))
0¹(L(2(x0))) → 2¹(2(R(x0)))
2¹(c(1(x0))) → 0¹(L(3(x0)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 2¹(c(0(x1))) → 0¹(0(x1)) at position [0] we obtained the following new rules:

2¹(c(0(L(x0)))) → 0¹(2(R(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

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

2¹(L(x1)) → 2¹(x1)
2¹(c(0(L(x0)))) → 0¹(2(R(x0)))
0¹(L(3(x0))) → 2¹(3(R(x0)))
0¹(L(2(x0))) → 2¹(2(R(x0)))
2¹(c(1(x0))) → 0¹(L(3(x0)))

The TRS R consists of the following rules:

R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))

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