The Runtime Complexity (full) 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(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
rev2(x, ++(y, z)) →⁺ ++(rev1(x, rev(rev2(y, z))), rev2(x, rev(rev2(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
The pumping substitution is [z / ++(y, z)].
The result substitution is [x / y].

The rewrite sequence
rev2(x, ++(y, z)) →⁺ ++(rev1(x, rev(rev2(y, z))), rev2(x, rev(rev2(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1,0].
The pumping substitution is [z / ++(y, z)].
The result substitution is [x / y].

(2) BOUNDS(2^n, 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:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

(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:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_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(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

Induction Base:
rev1(hole_rev12_0, gen_nil:++3_0(0)) →_R^Ω(1)
hole_rev12_0

Induction Step:
rev1(hole_rev12_0, gen_nil:++3_0(+(n5_0, 1))) →_R^Ω(1)
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) →_IH
hole_rev12_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:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Lemmas:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_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 (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
rev2(hole_rev12_0, gen_nil:++3_0(+(1, n116_0))) → *4_0, rt ∈ Ω(n116₀)

Induction Base:
rev2(hole_rev12_0, gen_nil:++3_0(+(1, 0)))

Induction Step:
rev2(hole_rev12_0, gen_nil:++3_0(+(1, +(n116_0, 1)))) →_R^Ω(1)
rev(++(hole_rev12_0, rev(rev2(hole_rev12_0, gen_nil:++3_0(+(1, n116_0)))))) →_IH
rev(++(hole_rev12_0, rev(*4_0)))

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

(13) Complex Obligation (BEST)

(14) Obligation:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Lemmas:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)
rev2(hole_rev12_0, gen_nil:++3_0(+(1, n116_0))) → *4_0, rt ∈ Ω(n116₀)

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_0(x))

The following defined symbols remain to be analysed:
rev

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

(15) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(16) Obligation:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Lemmas:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)
rev2(hole_rev12_0, gen_nil:++3_0(+(1, n116_0))) → *4_0, rt ∈ Ω(n116₀)

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_0(x))

No more defined symbols left to analyse.

(17) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

(18) BOUNDS(n^1, INF)

(19) Obligation:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Lemmas:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)
rev2(hole_rev12_0, gen_nil:++3_0(+(1, n116_0))) → *4_0, rt ∈ Ω(n116₀)

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_0(x))

No more defined symbols left to analyse.

(20) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

(21) BOUNDS(n^1, INF)

(22) Obligation:

TRS:
Rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Types:
rev :: nil:++ → nil:++
nil :: nil:++
++ :: rev1 → nil:++ → nil:++
rev1 :: rev1 → nil:++ → rev1
rev2 :: rev1 → nil:++ → nil:++
hole_nil:++1_0 :: nil:++
hole_rev12_0 :: rev1
gen_nil:++3_0 :: Nat → nil:++

Lemmas:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

Generator Equations:
gen_nil:++3_0(0) ⇔ nil
gen_nil:++3_0(+(x, 1)) ⇔ ++(hole_rev12_0, gen_nil:++3_0(x))

No more defined symbols left to analyse.

(23) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
rev1(hole_rev12_0, gen_nil:++3_0(n5_0)) → hole_rev12_0, rt ∈ Ω(1 + n5₀)

(24) BOUNDS(n^1, INF)