(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: SCNP Order with the following components:
Level mapping:
Top level AFS:
COPY(
x0,
x1,
x2,
x3) =
COPY(
x1)
Tags:
COPY has argument tags [1,1,0,2] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:
POL(COPY(x1, x2, x3)) = 1
POL(cons(x1, x2)) = x1
POL(copy(x1, x2, x3)) = 1 + x2 + x3
POL(f(x1)) = x1
POL(n) = 0
POL(nil) = 0
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
none
(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