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

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

c(c(y)) → y
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1, x2)) = x1 + x2   
POL(c(x1)) = 1 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 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 2 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

C(a(c(c(y)), x)) → C(c(c(a(x, 0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(c(y)), x)) → C(c(c(a(x, 0)))) at position [0] we obtained the following new rules:

C(a(c(c(y0)), a(x0, 0))) → C(c(c(x0)))
C(a(c(c(y0)), c(c(x0)))) → C(c(a(c(c(c(a(0, 0)))), x0)))



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

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

C(a(c(c(y0)), a(x0, 0))) → C(c(c(x0)))
C(a(c(c(y0)), c(c(x0)))) → C(c(a(c(c(c(a(0, 0)))), x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(c(y0)), c(c(x0)))) → C(c(a(c(c(c(a(0, 0)))), x0))) at position [0] we obtained the following new rules:

C(a(c(c(y0)), c(c(x1)))) → C(a(c(c(c(a(x1, 0)))), c(a(0, 0))))



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

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

C(a(c(c(y0)), c(c(x1)))) → C(a(c(c(c(a(x1, 0)))), c(a(0, 0))))
C(a(c(c(y0)), a(x0, 0))) → C(c(c(x0)))

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 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 1 less node.

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

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

C(a(c(c(y0)), a(x0, 0))) → C(c(c(x0)))

The TRS R consists of the following rules:

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

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