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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

B(b(c(x1))) → A(x1)


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
QDP
          ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

B(a(x1)) → A(b(x1))
A(a(x1)) → B(x1)
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 2 + 2·x1   
POL(B(x1)) = 10 + 2·x1   
POL(C(x1)) = 7 + 6·x1   
POL(a(x1)) = 4 + x1   
POL(b(x1)) = 8 + x1   
POL(c(x1)) = 3 + 3·x1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(a(x1)) → A(b(x1))
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
b(a(x1)) → a(b(x1))
b(b(c(x1))) → c(a(x1))
b(b(x1)) → a(a(a(x1)))
c(a(x1)) → b(a(c(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(a(x1)) → A(b(x1))
A(a(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 4·x1   
POL(B(x1)) = 7 + 4·x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 3 + x1   
POL(c(x1)) = 3·x1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
QDP
                  ↳ PisEmptyProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.