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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
A(y, 0) → B(y, 0)
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))
C(c(c(y))) → A(c(b(0, y)), 0)
C(c(c(y))) → B(0, y)
C(c(c(y))) → C(b(0, y))
C(c(c(y))) → A(a(c(b(0, y)), 0), 0)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
A(y, 0) → B(y, 0)
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))
C(c(c(y))) → A(c(b(0, y)), 0)
C(c(c(y))) → B(0, y)
C(c(c(y))) → C(b(0, y))
C(c(c(y))) → A(a(c(b(0, y)), 0), 0)

The TRS R consists of the following rules:

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

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 5 less nodes.

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

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

C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))

The TRS R consists of the following rules:

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

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(y))) → C(a(a(c(b(0, y)), 0), 0))
The remaining pairs can at least be oriented weakly.

C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
Used ordering: Polynomial interpretation [25,35]:

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

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



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

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

C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0)))

The TRS R consists of the following rules:

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

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(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (2)x_1   
POL(c(x1)) = 5/2 + (1/2)x_1   
POL(b(x1, x2)) = (1/4)x_1 + (4)x_2   
POL(a(x1, x2)) = (1/2)x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented:

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



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

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

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

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.