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:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

Q is empty.

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

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))

The set Q is empty.

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

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(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:

B(l(x1)) → A(r(x1))
B(l(x1)) → R(x1)
R(a(x1)) → R(x1)
R(a(x1)) → A(r(x1))
B(l(x1)) → B(a(r(x1)))
A(l(x1)) → A(x1)

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

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

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

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

B(l(x1)) → A(r(x1))
B(l(x1)) → R(x1)
R(a(x1)) → R(x1)
R(a(x1)) → A(r(x1))
B(l(x1)) → B(a(r(x1)))
A(l(x1)) → A(x1)

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(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 3 SCCs with 3 less nodes.

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

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

A(l(x1)) → A(x1)

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

A(l(x1)) → A(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

A(l(x1)) → A(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:

A(l(x1)) → A(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(l(x1)) = 2·x1   



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

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

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

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

R(a(x1)) → R(x1)

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP
  ↳ QTRS Reverse

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

R(a(x1)) → R(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ QDP
  ↳ QTRS Reverse

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

R(a(x1)) → R(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:

R(a(x1)) → R(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
  ↳ QTRS Reverse

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

B(l(x1)) → B(a(r(x1)))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

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

B(l(b(x0))) → B(a(l(b(x0))))
B(l(a(x0))) → B(a(a(r(x0))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
  ↳ QTRS Reverse

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

B(l(a(x0))) → B(a(a(r(x0))))
B(l(b(x0))) → B(a(l(b(x0))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(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.


B(l(b(x0))) → B(a(l(b(x0))))
The remaining pairs can at least be oriented weakly.

B(l(a(x0))) → B(a(a(r(x0))))
Used ordering: Polynomial Order [21,25] with Interpretation:

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


POL( b(x1) ) = x1 + 1


POL( l(x1) ) = x1


POL( B(x1) ) = x1


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



The following usable rules [17] were oriented:

a(l(x1)) → l(a(x1))



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

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

B(l(a(x0))) → B(a(a(r(x0))))

The TRS R consists of the following rules:

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPToSRSProof
                      ↳ QTRS
                        ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))
B(l(a(x0))) → B(a(a(r(x0))))

Q is empty.

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

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))
B(l(a(x0))) → B(a(a(r(x0))))

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

The set Q is empty.

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

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

Q is empty.

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

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

a(l(x)) → l(a(x))
r(a(x)) → a(r(x))
b(l(x)) → b(a(r(x)))
r(b(x)) → l(b(x))
B(l(a(x))) → B(a(a(r(x))))

The set Q is empty.

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

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

a(l(x)) → l(a(x))
r(a(x)) → a(r(x))
b(l(x)) → b(a(r(x)))
r(b(x)) → l(b(x))
B(l(a(x))) → B(a(a(r(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:

L(a(x)) → A(l(x))
L(a(x)) → L(x)
A(l(B(x))) → A(a(B(x)))
L(b(x)) → A(b(x))
B1(r(x)) → L(x)
B1(r(x)) → B1(l(x))
A(r(x)) → A(x)
A(l(B(x))) → A(B(x))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

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

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

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

L(a(x)) → A(l(x))
L(a(x)) → L(x)
A(l(B(x))) → A(a(B(x)))
L(b(x)) → A(b(x))
B1(r(x)) → L(x)
B1(r(x)) → B1(l(x))
A(r(x)) → A(x)
A(l(B(x))) → A(B(x))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(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 3 SCCs with 5 less nodes.

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

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

A(r(x)) → A(x)

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPToSRSProof
                      ↳ QTRS
                        ↳ QTRS Reverse
                          ↳ QTRS
                            ↳ QTRS Reverse
                            ↳ DependencyPairsProof
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ AND
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ UsableRulesReductionPairsProof
                                      ↳ UsableRulesProof
                                    ↳ QDP
                                    ↳ QDP
                            ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A(r(x)) → A(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:

A(r(x)) → A(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(r(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPToSRSProof
                      ↳ QTRS
                        ↳ QTRS Reverse
                          ↳ QTRS
                            ↳ QTRS Reverse
                            ↳ DependencyPairsProof
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ AND
                                    ↳ 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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPToSRSProof
                      ↳ QTRS
                        ↳ QTRS Reverse
                          ↳ QTRS
                            ↳ QTRS Reverse
                            ↳ 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:

A(r(x)) → A(x)

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

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

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

L(a(x)) → L(x)

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ Narrowing
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPToSRSProof
                      ↳ QTRS
                        ↳ QTRS Reverse
                          ↳ QTRS
                            ↳ QTRS Reverse
                            ↳ DependencyPairsProof
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ AND
                                    ↳ QDP
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ UsableRulesReductionPairsProof
                                      ↳ UsableRulesProof
                                    ↳ QDP
                            ↳ QTRS Reverse
  ↳ QTRS Reverse

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

L(a(x)) → L(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:

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

Used ordering: POLO with Polynomial interpretation [25]:

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



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

L(a(x)) → L(x)

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

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

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

B1(r(x)) → B1(l(x))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(r(x)) → B1(l(x)) at position [0] we obtained the following new rules:

B1(r(a(x0))) → B1(a(l(x0)))
B1(r(b(x0))) → B1(r(a(b(x0))))



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

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

B1(r(a(x0))) → B1(a(l(x0)))
B1(r(b(x0))) → B1(r(a(b(x0))))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(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.


B1(r(b(x0))) → B1(r(a(b(x0))))
The remaining pairs can at least be oriented weakly.

B1(r(a(x0))) → B1(a(l(x0)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( B1(x1) ) = x1


POL( r(x1) ) = x1


POL( b(x1) ) = x1 + 1


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


POL( B(x1) ) = 1


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



The following usable rules [17] were oriented:

a(r(x)) → r(a(x))
a(l(B(x))) → r(a(a(B(x))))



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

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

B1(r(a(x0))) → B1(a(l(x0)))

The TRS R consists of the following rules:

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))
a(l(B(x))) → r(a(a(B(x))))

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

a(l(x)) → l(a(x))
r(a(x)) → a(r(x))
b(l(x)) → b(a(r(x)))
r(b(x)) → l(b(x))
B(l(a(x))) → B(a(a(r(x))))

The set Q is empty.

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

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

a(l(x)) → l(a(x))
r(a(x)) → a(r(x))
b(l(x)) → b(a(r(x)))
r(b(x)) → l(b(x))
B(l(a(x))) → B(a(a(r(x))))

Q is empty.

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

a(l(x1)) → l(a(x1))
r(a(x1)) → a(r(x1))
b(l(x1)) → b(a(r(x1)))
r(b(x1)) → l(b(x1))

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))

The set Q is empty.

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

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

l(a(x)) → a(l(x))
a(r(x)) → r(a(x))
l(b(x)) → r(a(b(x)))
b(r(x)) → b(l(x))

Q is empty.