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
          ↳ RuleRemovalProof

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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(C(x1)) = 2·x1   
POL(a(x1, x2)) = x1 + x2   
POL(b(x1, x2)) = x1 + x2   
POL(c(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing

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.
By narrowing [15] the rule C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0))) at position [0] we obtained the following new rules:

C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing

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

C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 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.
By narrowing [15] the rule C(c(c(y0))) → C(c(b(a(c(b(0, y0)), 0), 0))) at position [0] we obtained the following new rules:

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof

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

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))
C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing

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

C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 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.
By narrowing [15] the rule C(c(c(y0))) → C(c(a(b(c(b(0, y0)), 0), 0))) at position [0] we obtained the following new rules:

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 0), 0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof

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

C(c(c(y0))) → C(c(b(b(c(b(0, y0)), 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 0 SCCs with 1 less node.