Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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))
COPY(s(x), y, z) → F(y)

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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
COPY(x1, x2, x3)  =  COPY(x1, x2, x3)
s(x1)  =  s(x1)
cons(x1, x2)  =  x2
f(x1)  =  f(x1)
nil  =  nil
copy(x1, x2, x3)  =  copy(x3)
n  =  n
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:

[COPY₃, s₁] > [f₁, copy₁, 0] > n
nil > [f₁, copy₁, 0] > n

Status:

COPY₃: [1,2,3]
s₁: [1]
f₁: [1]
nil: []
copy₁: [1]
n: []
0: []

The following usable rules [FROCOS05] were oriented:

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))

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE