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(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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


A(b(c(x1))) → A(a(b(x1)))
A(b(c(x1))) → A(b(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 1/4 + (4)x_1   
POL(a(x1)) = (4)x_1   
POL(A(x1)) = (1/4)x_1   
POL(b(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 1/64.
The following usable rules [17] were oriented:

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



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

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

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

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

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

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

Q is empty.