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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

A(c(x1)) → B(x1)


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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


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

A(x1) → B(b(b(x1)))
B(c(b(x1))) → A(c(x1))
Used ordering: Polynomial interpretation [25,35]:

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

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



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

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

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

The TRS R consists of the following rules:

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

A(x1) → B(b(b(x1)))
Used ordering: Polynomial interpretation [25,35]:

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

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



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

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

A(x1) → B(b(b(x1)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
We have reversed the following QTRS:
The set of rules R is

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.