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:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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:

F(x1) → C(c(x1))
C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
N(f(x1)) → F(n(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(s(x1)) → F(s(s(x1)))
N(f(x1)) → N(x1)
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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:

F(x1) → C(c(x1))
C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
N(f(x1)) → F(n(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(s(x1)) → F(s(s(x1)))
N(f(x1)) → N(x1)
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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.


N(s(x1)) → F(s(s(x1)))
The remaining pairs can at least be oriented weakly.

F(x1) → C(c(x1))
C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
N(f(x1)) → F(n(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(f(x1)) → N(x1)
F(x1) → N(c(c(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 0   
POL(N(x1)) = (1/4)x_1   
POL(c(x1)) = 0   
POL(f(x1)) = (2)x_1   
POL(n(x1)) = (2)x_1   
POL(s(x1)) = 1   
POL(F(x1)) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(s(x1)))
f(x1) → n(c(c(x1)))



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

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

F(x1) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
C(f(x1)) → C(c(x1))
C(f(x1)) → C(x1)
N(f(x1)) → F(n(x1))
F(x1) → C(x1)
F(x1) → N(c(c(x1)))
N(f(x1)) → N(x1)

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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.


F(x1) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
C(f(x1)) → C(c(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(f(x1)) → N(x1)
The remaining pairs can at least be oriented weakly.

N(f(x1)) → F(n(x1))
F(x1) → N(c(c(x1)))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 1/4 + (5/2)x_1   
POL(N(x1)) = 1/2 + (5/2)x_1   
POL(c(x1)) = x_1   
POL(f(x1)) = 13/4 + (13/4)x_1   
POL(n(x1)) = 13/4 + (13/4)x_1   
POL(s(x1)) = 0   
POL(F(x1)) = 1/2 + (5/2)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(s(x1)))
f(x1) → n(c(c(x1)))



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

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

N(f(x1)) → F(n(x1))
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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