(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)
s(
x1) =
s(
x1)
cons(
x1,
x2) =
x2
f(
x1) =
x1
nil =
nil
copy(
x1,
x2,
x3) =
x3
n =
n
0 =
0
Recursive Path Order [RPO].
Precedence:
s1 > COPY1
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