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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x1) → x1
b(x1) → a(x1)
b(a(c(x1))) → c(c(b(a(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.