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:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → 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:

C(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → B(x)
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

C(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → B(x)
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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.


C(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
A(1, x) → B(x)
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)
The remaining pairs can at least be oriented weakly.

C(c(c(b(x)))) → A(1, b(c(x)))
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (1/4)x_1   
POL(c(x1)) = 1 + (2)x_1   
POL(B(x1)) = (1/4)x_1   
POL(a(x1, x2)) = 3 + (4)x_2   
POL(A(x1, x2)) = 1/4 + (1/2)x_2   
POL(b(x1)) = 1 + (2)x_1   
POL(1) = 0   
POL(0) = 5/4   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
c(c(c(b(x)))) → a(1, b(c(x)))



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

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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.


C(c(c(b(x)))) → A(1, b(c(x)))
The remaining pairs can at least be oriented weakly.

A(1, x) → C(b(x))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 1/4 + (1/2)x_1   
POL(c(x1)) = 1/2 + (3/2)x_1   
POL(a(x1, x2)) = 1/2 + (1/4)x_1 + (9/4)x_2   
POL(A(x1, x2)) = 1/4 + (3)x_1 + (3/4)x_2   
POL(b(x1)) = (3/2)x_1   
POL(1) = 0   
POL(0) = 3   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
c(c(c(b(x)))) → a(1, b(c(x)))



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

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

A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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.