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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(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:

half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(4) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

(5) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
half, lastbit, conv

They will be analysed ascendingly in the following order:
half < conv
lastbit < conv

(6) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

The following defined symbols remain to be analysed:
half, lastbit, conv

They will be analysed ascendingly in the following order:
half < conv
lastbit < conv

(7) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

Induction Base:
half(gen_0':s3_0(*(2, 0))) →_R^Ω(1)
0'

Induction Step:
half(gen_0':s3_0(*(2, +(n6_0, 1)))) →_R^Ω(1)
s(half(gen_0':s3_0(*(2, n6_0)))) →_IH
s(gen_0':s3_0(c7_0))

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

(8) Complex Obligation (BEST)

(9) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

The following defined symbols remain to be analysed:
lastbit, conv

They will be analysed ascendingly in the following order:
lastbit < conv

(10) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
lastbit(gen_0':s3_0(*(2, n300_0))) → gen_0':s3_0(0), rt ∈ Ω(1 + n300₀)

Induction Base:
lastbit(gen_0':s3_0(*(2, 0))) →_R^Ω(1)
0'

Induction Step:
lastbit(gen_0':s3_0(*(2, +(n300_0, 1)))) →_R^Ω(1)
lastbit(gen_0':s3_0(*(2, n300_0))) →_IH
gen_0':s3_0(0)

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

(11) Complex Obligation (BEST)

(12) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)
lastbit(gen_0':s3_0(*(2, n300_0))) → gen_0':s3_0(0), rt ∈ Ω(1 + n300₀)

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

The following defined symbols remain to be analysed:
conv

(13) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(14) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)
lastbit(gen_0':s3_0(*(2, n300_0))) → gen_0':s3_0(0), rt ∈ Ω(1 + n300₀)

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

No more defined symbols left to analyse.

(15) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

(16) BOUNDS(n^1, INF)

(17) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)
lastbit(gen_0':s3_0(*(2, n300_0))) → gen_0':s3_0(0), rt ∈ Ω(1 + n300₀)

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

No more defined symbols left to analyse.

(18) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

(19) BOUNDS(n^1, INF)

(20) Obligation:

TRS:
Rules:
half(0') → 0'
half(s(0')) → 0'
half(s(s(x))) → s(half(x))
lastbit(0') → 0'
lastbit(s(0')) → s(0')
lastbit(s(s(x))) → lastbit(x)
conv(0') → cons(nil, 0')
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

Types:
half :: 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
lastbit :: 0':s → 0':s
conv :: 0':s → nil:cons
cons :: nil:cons → 0':s → nil:cons
nil :: nil:cons
hole_0':s1_0 :: 0':s
hole_nil:cons2_0 :: nil:cons
gen_0':s3_0 :: Nat → 0':s
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_0':s3_0(0) ⇔ 0'
gen_0':s3_0(+(x, 1)) ⇔ s(gen_0':s3_0(x))
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(gen_nil:cons4_0(x), 0')

No more defined symbols left to analyse.

(21) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
half(gen_0':s3_0(*(2, n6_0))) → gen_0':s3_0(n6_0), rt ∈ Ω(1 + n6₀)

(22) BOUNDS(n^1, INF)