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:

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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:

COPY(s(x), y, z) → COPY(x, y, cons(f(y), z))
F(cons(f(cons(nil, y)), z)) → COPY(n, y, z)
COPY(s(x), y, z) → F(y)
COPY(0, y, z) → F(z)

The TRS R consists of the following rules:

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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:

COPY(s(x), y, z) → COPY(x, y, cons(f(y), z))
F(cons(f(cons(nil, y)), z)) → COPY(n, y, z)
COPY(s(x), y, z) → F(y)
COPY(0, y, z) → F(z)

The TRS R consists of the following rules:

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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:

COPY(s(x), y, z) → COPY(x, y, cons(f(y), z))

The TRS R consists of the following rules:

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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.


COPY(s(x), y, z) → COPY(x, y, cons(f(y), z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1 + (3)x_2   
POL(n) = 0   
POL(f(x1)) = 3   
POL(COPY(x1, x2, x3)) = (3)x_1   
POL(s(x1)) = 3 + (4)x_1   
POL(nil) = 2   
POL(copy(x1, x2, x3)) = 3 + (2)x_2 + (4)x_3   
The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented: none



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

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

f(cons(nil, y)) → y
f(cons(f(cons(nil, y)), z)) → copy(n, y, z)
copy(0, y, z) → f(z)
copy(s(x), y, z) → copy(x, y, cons(f(y), z))

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.