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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

Rewrite Strategy: FULL

(1) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

first(0', X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
cons/0
from/0

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

Types:
first :: 0':s → nil:cons → nil:cons
0' :: 0':s
nil :: nil:cons
s :: 0':s → 0':s
cons :: nil:cons → nil:cons
from :: nil:cons
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_0':s4_0 :: Nat → 0':s

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

Heuristically decided to analyse the following defined symbols:
first, from

(8) Obligation:

TRS:
Rules:
first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

Types:
first :: 0':s → nil:cons → nil:cons
0' :: 0':s
nil :: nil:cons
s :: 0':s → 0':s
cons :: nil:cons → nil:cons
from :: nil:cons
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_0':s4_0 :: Nat → 0':s

Generator Equations:
gen_nil:cons3_0(0) ⇔ nil
gen_nil:cons3_0(+(x, 1)) ⇔ cons(gen_nil:cons3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

The following defined symbols remain to be analysed:
first, from

(9) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

Induction Base:
first(gen_0':s4_0(0), gen_nil:cons3_0(0)) →_R^Ω(1)
nil

Induction Step:
first(gen_0':s4_0(+(n6_0, 1)), gen_nil:cons3_0(+(n6_0, 1))) →_R^Ω(1)
cons(first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0))) →_IH
cons(gen_nil:cons3_0(c7_0))

We have rt ∈ Ω(n¹) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(n).

(10) Complex Obligation (BEST)

(11) Obligation:

TRS:
Rules:
first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

Types:
first :: 0':s → nil:cons → nil:cons
0' :: 0':s
nil :: nil:cons
s :: 0':s → 0':s
cons :: nil:cons → nil:cons
from :: nil:cons
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_0':s4_0 :: Nat → 0':s

Lemmas:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons3_0(0) ⇔ nil
gen_nil:cons3_0(+(x, 1)) ⇔ cons(gen_nil:cons3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

The following defined symbols remain to be analysed:
from

(12) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

Could not prove a rewrite lemma for the defined symbol from.

(13) Obligation:

TRS:
Rules:
first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

Types:
first :: 0':s → nil:cons → nil:cons
0' :: 0':s
nil :: nil:cons
s :: 0':s → 0':s
cons :: nil:cons → nil:cons
from :: nil:cons
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_0':s4_0 :: Nat → 0':s

Lemmas:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons3_0(0) ⇔ nil
gen_nil:cons3_0(+(x, 1)) ⇔ cons(gen_nil:cons3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

No more defined symbols left to analyse.

(14) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

(15) BOUNDS(n^1, INF)

(16) Obligation:

TRS:
Rules:
first(0', X) → nil
first(s(X), cons(Z)) → cons(first(X, Z))
from → cons(from)

Types:
first :: 0':s → nil:cons → nil:cons
0' :: 0':s
nil :: nil:cons
s :: 0':s → 0':s
cons :: nil:cons → nil:cons
from :: nil:cons
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons
gen_0':s4_0 :: Nat → 0':s

Lemmas:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons3_0(0) ⇔ nil
gen_nil:cons3_0(+(x, 1)) ⇔ cons(gen_nil:cons3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

No more defined symbols left to analyse.

(17) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
first(gen_0':s4_0(n6_0), gen_nil:cons3_0(n6_0)) → gen_nil:cons3_0(n6_0), rt ∈ Ω(1 + n6₀)

(18) BOUNDS(n^1, INF)