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:

app(x, y) → helpa(0, plus(length(x), length(y)), x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0, cons(x, xs), ys) → x
take(0, nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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:

app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(4) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

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

Heuristically decided to analyse the following defined symbols:
helpa, plus, length, ge, helpb, take

They will be analysed ascendingly in the following order:
ge < helpa
helpa = helpb
take < helpb

(6) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
plus, helpa, length, ge, helpb, take

They will be analysed ascendingly in the following order:
ge < helpa
helpa = helpb
take < helpb

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

Proved the following rewrite lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

Induction Base:
plus(gen_0':s6_0(a), gen_0':s6_0(0)) →_R^Ω(1)
gen_0':s6_0(a)

Induction Step:
plus(gen_0':s6_0(a), gen_0':s6_0(+(n8_0, 1))) →_R^Ω(1)
s(plus(gen_0':s6_0(a), gen_0':s6_0(n8_0))) →_IH
s(gen_0':s6_0(+(a, c9_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:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
length, helpa, ge, helpb, take

They will be analysed ascendingly in the following order:
ge < helpa
helpa = helpb
take < helpb

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

Proved the following rewrite lemma:
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)

Induction Base:
length(gen_nil:cons:xs5_0(0)) →_R^Ω(1)
0'

Induction Step:
length(gen_nil:cons:xs5_0(+(n825_0, 1))) →_R^Ω(1)
s(length(gen_nil:cons:xs5_0(n825_0))) →_IH
s(gen_0':s6_0(c826_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:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
ge, helpa, helpb, take

They will be analysed ascendingly in the following order:
ge < helpa
helpa = helpb
take < helpb

(13) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)

Induction Base:
ge(gen_0':s6_0(0), gen_0':s6_0(0)) →_R^Ω(1)
true

Induction Step:
ge(gen_0':s6_0(+(n1099_0, 1)), gen_0':s6_0(+(n1099_0, 1))) →_R^Ω(1)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) →_IH
true

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

(14) Complex Obligation (BEST)

(15) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
take, helpa, helpb

They will be analysed ascendingly in the following order:
helpa = helpb
take < helpb

(16) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
take(gen_0':s6_0(n1422_0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, n1422_0))) → hole_take3_0, rt ∈ Ω(1 + n1422₀)

Induction Base:
take(gen_0':s6_0(0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, 0))) →_R^Ω(1)
hole_take3_0

Induction Step:
take(gen_0':s6_0(+(n1422_0, 1)), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, +(n1422_0, 1)))) →_R^Ω(1)
take(gen_0':s6_0(n1422_0), nil, gen_nil:cons:xs5_0(+(1, n1422_0))) →_IH
hole_take3_0

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

(17) Complex Obligation (BEST)

(18) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)
take(gen_0':s6_0(n1422_0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, n1422_0))) → hole_take3_0, rt ∈ Ω(1 + n1422₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
helpb, helpa

They will be analysed ascendingly in the following order:
helpa = helpb

(19) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(20) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)
take(gen_0':s6_0(n1422_0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, n1422_0))) → hole_take3_0, rt ∈ Ω(1 + n1422₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

The following defined symbols remain to be analysed:
helpa

They will be analysed ascendingly in the following order:
helpa = helpb

(21) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(22) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)
take(gen_0':s6_0(n1422_0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, n1422_0))) → hole_take3_0, rt ∈ Ω(1 + n1422₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

No more defined symbols left to analyse.

(23) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

(24) BOUNDS(n^1, INF)

(25) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)
take(gen_0':s6_0(n1422_0), gen_nil:cons:xs5_0(0), gen_nil:cons:xs5_0(+(1, n1422_0))) → hole_take3_0, rt ∈ Ω(1 + n1422₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

No more defined symbols left to analyse.

(26) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

(27) BOUNDS(n^1, INF)

(28) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)
ge(gen_0':s6_0(n1099_0), gen_0':s6_0(n1099_0)) → true, rt ∈ Ω(1 + n1099₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

No more defined symbols left to analyse.

(29) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

(30) BOUNDS(n^1, INF)

(31) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)
length(gen_nil:cons:xs5_0(n825_0)) → gen_0':s6_0(n825_0), rt ∈ Ω(1 + n825₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

No more defined symbols left to analyse.

(32) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

(33) BOUNDS(n^1, INF)

(34) Obligation:

TRS:
Rules:
app(x, y) → helpa(0', plus(length(x), length(y)), x, y)
plus(x, 0') → x
plus(x, s(y)) → s(plus(x, y))
length(nil) → 0'
length(cons(x, y)) → s(length(y))
helpa(c, l, ys, zs) → if(ge(c, l), c, l, ys, zs)
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(true, c, l, ys, zs) → nil
if(false, c, l, ys, zs) → helpb(c, l, ys, zs)
take(0', cons(x, xs), ys) → x
take(0', nil, cons(y, ys)) → y
take(s(c), cons(x, xs), ys) → take(c, xs, ys)
take(s(c), nil, cons(y, ys)) → take(c, nil, ys)
helpb(c, l, ys, zs) → cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

Types:
app :: nil:cons:xs → nil:cons:xs → nil:cons:xs
helpa :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
0' :: 0':s
plus :: 0':s → 0':s → 0':s
length :: nil:cons:xs → 0':s
s :: 0':s → 0':s
nil :: nil:cons:xs
cons :: take → nil:cons:xs → nil:cons:xs
if :: true:false → 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
helpb :: 0':s → 0':s → nil:cons:xs → nil:cons:xs → nil:cons:xs
take :: 0':s → nil:cons:xs → nil:cons:xs → take
xs :: nil:cons:xs
hole_nil:cons:xs1_0 :: nil:cons:xs
hole_0':s2_0 :: 0':s
hole_take3_0 :: take
hole_true:false4_0 :: true:false
gen_nil:cons:xs5_0 :: Nat → nil:cons:xs
gen_0':s6_0 :: Nat → 0':s

Lemmas:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

Generator Equations:
gen_nil:cons:xs5_0(0) ⇔ nil
gen_nil:cons:xs5_0(+(x, 1)) ⇔ cons(hole_take3_0, gen_nil:cons:xs5_0(x))
gen_0':s6_0(0) ⇔ 0'
gen_0':s6_0(+(x, 1)) ⇔ s(gen_0':s6_0(x))

No more defined symbols left to analyse.

(35) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
plus(gen_0':s6_0(a), gen_0':s6_0(n8_0)) → gen_0':s6_0(+(n8_0, a)), rt ∈ Ω(1 + n8₀)

(36) BOUNDS(n^1, INF)