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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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

POL(C(x1)) = 1 + x_1   
POL(c(x1)) = 1/2   
POL(b(x1, x2)) = x_1 + x_2   
POL(a(x1, x2)) = 1/2 + x_1   
POL(A(x1, x2)) = 5/4 + (7/4)x_1 + (2)x_2   
POL(0) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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