The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
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))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
copy(s(x), y, z) →⁺ copy(x, y, cons(f(y), z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x)].
The result substitution is [z / cons(f(y), z)].

(2) BOUNDS(n^1, INF)

(3) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(4) Obligation:

Runtime Complexity Relative TRS:
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))

S is empty.
Rewrite Strategy: FULL

(5) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(6) Obligation:

TRS:
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))

Types:
f :: nil:cons → nil:cons
cons :: nil:cons → nil:cons → nil:cons
nil :: nil:cons
copy :: n:0':s → nil:cons → nil:cons → nil:cons
n :: n:0':s
0' :: n:0':s
s :: n:0':s → n:0':s
hole_nil:cons1_0 :: nil:cons
hole_n:0':s2_0 :: n:0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_n:0':s4_0 :: Nat → n:0':s

(7) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
f, copy

They will be analysed ascendingly in the following order:
f = copy

(8) Obligation:

TRS:
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))

Types:
f :: nil:cons → nil:cons
cons :: nil:cons → nil:cons → nil:cons
nil :: nil:cons
copy :: n:0':s → nil:cons → nil:cons → nil:cons
n :: n:0':s
0' :: n:0':s
s :: n:0':s → n:0':s
hole_nil:cons1_0 :: nil:cons
hole_n:0':s2_0 :: n:0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_n:0':s4_0 :: Nat → n:0':s

Generator Equations:
gen_nil:cons3_0(0) ⇔ nil
gen_nil:cons3_0(+(x, 1)) ⇔ cons(nil, gen_nil:cons3_0(x))
gen_n:0':s4_0(0) ⇔ 0'
gen_n:0':s4_0(+(x, 1)) ⇔ s(gen_n:0':s4_0(x))

The following defined symbols remain to be analysed:
copy, f

They will be analysed ascendingly in the following order:
f = copy