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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → c(c(x1))
b(c(x1)) → a(a(b(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

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

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

B(c(c(x0))) → B(a(a(b(b(x0)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof

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

B(c(c(x0))) → B(a(a(b(b(x0)))))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

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

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

a(b(x1)) → x1
a(c(x1)) → c(c(x1))
b(c(x1)) → a(a(b(b(x1))))
B(c(c(x0))) → B(a(a(b(b(x0)))))
B(c(x1)) → B(x1)

Q is empty.

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

a(b(x1)) → x1
a(c(x1)) → c(c(x1))
b(c(x1)) → a(a(b(b(x1))))
B(c(c(x0))) → B(a(a(b(b(x0)))))
B(c(x1)) → B(x1)

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

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

The set Q is empty.

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

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

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

C(b(x)) → B1(b(a(a(x))))
C(a(x)) → C(c(x))
C(a(x)) → C(x)
C(c(B(x))) → B1(b(a(a(B(x)))))
C(c(B(x))) → B1(a(a(B(x))))
C(b(x)) → B1(a(a(x)))

The TRS R consists of the following rules:

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

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

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

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

C(b(x)) → B1(b(a(a(x))))
C(a(x)) → C(c(x))
C(a(x)) → C(x)
C(c(B(x))) → B1(b(a(a(B(x)))))
C(c(B(x))) → B1(a(a(B(x))))
C(b(x)) → B1(a(a(x)))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → c(c(x))
c(b(x)) → b(b(a(a(x))))
c(c(B(x))) → b(b(a(a(B(x)))))
c(B(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 1 SCC with 4 less nodes.

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

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

C(a(x)) → C(c(x))
C(a(x)) → C(x)

The TRS R consists of the following rules:

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


C(a(x)) → C(x)
The remaining pairs can at least be oriented weakly.

C(a(x)) → C(c(x))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x1) ) = x1 + 1


POL( c(x1) ) = x1 + 1


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


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


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

C(a(x)) → C(c(x))

The TRS R consists of the following rules:

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


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

POL(C(x1)) = x_1   
POL(c(x1)) = 15/4 + (2)x_1   
POL(B(x1)) = 15/4 + (3/4)x_1   
POL(a(x1)) = 4 + (2)x_1   
POL(b(x1)) = 5/4 + (1/2)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

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