The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Rewrite Strategy: INNERMOST

(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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

S is empty.
Rewrite Strategy: INNERMOST

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

Sliced the following arguments:
s/0

(4) Obligation:

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s, nil) → s
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

S is empty.
Rewrite Strategy: INNERMOST

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

Infered types.

(6) Obligation:

Innermost TRS:
Rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s, nil) → s
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Types:
rev :: nil:cons → nil:cons
nil :: nil:cons
cons :: 0':s → nil:cons → nil:cons
rev1 :: 0':s → nil:cons → 0':s
rev2 :: 0':s → nil:cons → nil:cons
0' :: 0':s
s :: 0':s
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons

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

Heuristically decided to analyse the following defined symbols:
rev, rev1, rev2

They will be analysed ascendingly in the following order:
rev1 < rev
rev = rev2

(8) Obligation:

Innermost TRS:
Rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s, nil) → s
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Types:
rev :: nil:cons → nil:cons
nil :: nil:cons
cons :: 0':s → nil:cons → nil:cons
rev1 :: 0':s → nil:cons → 0':s
rev2 :: 0':s → nil:cons → nil:cons
0' :: 0':s
s :: 0':s
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons

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

The following defined symbols remain to be analysed:
rev1, rev, rev2

They will be analysed ascendingly in the following order:
rev1 < rev
rev = rev2

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

Proved the following rewrite lemma:
rev1(0', gen_nil:cons3_0(n5_0)) → 0', rt ∈ Ω(1 + n5₀)

Induction Base:
rev1(0', gen_nil:cons3_0(0)) →_R^Ω(1)
0'

Induction Step:
rev1(0', gen_nil:cons3_0(+(n5_0, 1))) →_R^Ω(1)
rev1(0', gen_nil:cons3_0(n5_0)) →_IH
0'

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

(10) Complex Obligation (BEST)

(11) Obligation:

Innermost TRS:
Rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s, nil) → s
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Types:
rev :: nil:cons → nil:cons
nil :: nil:cons
cons :: 0':s → nil:cons → nil:cons
rev1 :: 0':s → nil:cons → 0':s
rev2 :: 0':s → nil:cons → nil:cons
0' :: 0':s
s :: 0':s
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons

Lemmas:
rev1(0', gen_nil:cons3_0(n5_0)) → 0', rt ∈ Ω(1 + n5₀)

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

The following defined symbols remain to be analysed:
rev2, rev

They will be analysed ascendingly in the following order:
rev = rev2

(12) RewriteLemmaProof (EQUIVALENT transformation)

Proved the following rewrite lemma:
rev2(0', gen_nil:cons3_0(n215_0)) → gen_nil:cons3_0(n215_0), rt ∈ Ω(2ⁿ)

Induction Base:
rev2(0', gen_nil:cons3_0(0)) →_R^Ω(1)
nil

Induction Step:
rev2(0', gen_nil:cons3_0(+(n215_0, 1))) →_R^Ω(1)
rev(cons(0', rev2(0', gen_nil:cons3_0(n215_0)))) →_IH
rev(cons(0', gen_nil:cons3_0(c216_0))) →_R^Ω(1)
cons(rev1(0', gen_nil:cons3_0(n215_0)), rev2(0', gen_nil:cons3_0(n215_0))) →_L^{Ω(1 + n215₀)}
cons(0', rev2(0', gen_nil:cons3_0(n215_0))) →_IH
cons(0', gen_nil:cons3_0(c216_0))

We have rt ∈ Ω(2ⁿ) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(2ⁿ)

(13) BOUNDS(2^n, INF)

(14) Obligation:

Innermost TRS:
Rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0', nil) → 0'
rev1(s, nil) → s
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Types:
rev :: nil:cons → nil:cons
nil :: nil:cons
cons :: 0':s → nil:cons → nil:cons
rev1 :: 0':s → nil:cons → 0':s
rev2 :: 0':s → nil:cons → nil:cons
0' :: 0':s
s :: 0':s
hole_nil:cons1_0 :: nil:cons
hole_0':s2_0 :: 0':s
gen_nil:cons3_0 :: Nat → nil:cons

Lemmas:
rev1(0', gen_nil:cons3_0(n5_0)) → 0', rt ∈ Ω(1 + n5₀)

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

No more defined symbols left to analyse.

(15) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
rev1(0', gen_nil:cons3_0(n5_0)) → 0', rt ∈ Ω(1 + n5₀)

(16) BOUNDS(n^1, INF)