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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

A(c(x1)) → A(b(c(a(x1))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = x1 + 1


POL( c(x1) ) = x1 + 1


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


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

A(c(x1)) → A(b(c(a(x1))))

The TRS R consists of the following rules:

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

POL(c(x1)) = 1/2 + (4)x_1   
POL(a(x1)) = (4)x_1   
POL(A(x1)) = (1/2)x_1   
POL(b(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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(c(x1)) → b(c(a(b(c(a(x1))))))
b(c(x1)) → 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(c(x1)) → b(c(a(b(c(a(x1))))))
b(c(x1)) → x1

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.