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

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

Q is empty.

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

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

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))

Q is empty.

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

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

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

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

A(b(x1)) → A(a(a(x1)))
B(a(x1)) → A(a(x1))
A(b(x1)) → B(a(a(a(x1))))
A(a(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(x1))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

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

A(b(x1)) → A(a(a(x1)))
B(a(x1)) → A(a(x1))
A(b(x1)) → B(a(a(a(x1))))
A(a(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(x1))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

A(b(x1)) → A(a(a(x1)))
B(a(x1)) → A(a(x1))
A(b(x1)) → B(a(a(a(x1))))
A(b(x1)) → A(a(x1))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(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.


A(b(x1)) → A(a(a(x1)))
A(b(x1)) → B(a(a(a(x1))))
A(b(x1)) → A(a(x1))
A(b(x1)) → A(x1)
The remaining pairs can at least be oriented weakly.

B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = x1 + 1


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


POL( b(x1) ) = x1 + 1


POL( B(x1) ) = x1 + 1


POL( a(x1) ) = x1



The following usable rules [17] were oriented:

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ QDPToSRSProof

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

B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
QTRS
                  ↳ QTRS Reverse

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))
B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)

Q is empty.

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

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))
B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

Q is empty.

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

Q is empty.

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
QTRS
                      ↳ DependencyPairsProof

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(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:

A1(B(x)) → A1(A(x))
B1(a(x)) → B1(x)
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
A1(a(x)) → B1(c(a(x)))
B1(a(x)) → A1(a(a(b(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
QDP
                          ↳ DependencyGraphProof

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

A1(B(x)) → A1(A(x))
B1(a(x)) → B1(x)
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
A1(a(x)) → B1(c(a(x)))
B1(a(x)) → A1(a(a(b(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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 5 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
                              ↳ QDP

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

A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ 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:

A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
QDP
                                    ↳ RuleRemovalProof
                              ↳ QDP

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

A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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:

a(A(x)) → B(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + 2·x1   
POL(A1(x1)) = 2·x1   
POL(B(x1)) = 1 + 2·x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = 2·x1   
POL(c(x1)) = x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
QDP
                                        ↳ UsableRulesReductionPairsProof
                              ↳ QDP

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

A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))

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:

a(B(x)) → a(A(x))
Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = x1   
POL(B(x1)) = 2 + x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = 2·x1   
POL(c(x1)) = x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ UsableRulesReductionPairsProof
QDP
                                            ↳ RuleRemovalProof
                              ↳ QDP

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

A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))

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:

A1(b(x)) → A1(x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A1(x1)) = 2·x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ UsableRulesReductionPairsProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof
QDP
                                                ↳ QDPOrderProof
                              ↳ QDP

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

A1(b(x)) → A1(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))

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.


A1(b(x)) → A1(a(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 0   
POL(A1(x1)) = x_1   
POL(a(x1)) = 2 + (2)x_1   
POL(b(x1)) = 3 + (4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

a(a(x)) → b(c(a(x)))
a(b(x)) → a(a(x))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
                                      ↳ QDP
                                        ↳ UsableRulesReductionPairsProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof
                                              ↳ QDP
                                                ↳ QDPOrderProof
QDP
                                                    ↳ PisEmptyProof
                              ↳ QDP

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

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))

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
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof

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

B1(a(x)) → B1(x)

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ UsableRulesReductionPairsProof
                                ↳ UsableRulesProof

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

B1(a(x)) → B1(x)

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:

B1(a(x)) → B1(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(B1(x1)) = 2·x1   
POL(a(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ UsableRulesReductionPairsProof
QDP
                                        ↳ PisEmptyProof
                                ↳ UsableRulesProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                ↳ UsableRulesProof
QDP

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

B1(a(x)) → B1(x)

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