(0) Obligation:

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
p(s(s(x))) →+ s(p(s(x)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / s(x)].
The result substitution is [ ].

(2) BOUNDS(n^1, 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:

minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

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

Heuristically decided to analyse the following defined symbols:
minus, p, div, plus

They will be analysed ascendingly in the following order:
p < minus
minus < div
minus < plus
plus < div

(8) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

The following defined symbols remain to be analysed:
p, minus, div, plus

They will be analysed ascendingly in the following order:
p < minus
minus < div
minus < plus
plus < div

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

Proved the following rewrite lemma:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

Induction Base:
p(gen_0':s2_0(+(2, 0)))

Induction Step:
p(gen_0':s2_0(+(2, +(n4_0, 1)))) →RΩ(1)
s(p(s(gen_0':s2_0(+(1, n4_0))))) →IH
s(*3_0)

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

(10) Complex Obligation (BEST)

(11) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

The following defined symbols remain to be analysed:
minus, div, plus

They will be analysed ascendingly in the following order:
minus < div
minus < plus
plus < div

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

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

(13) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

The following defined symbols remain to be analysed:
plus, div

They will be analysed ascendingly in the following order:
plus < div

(14) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(15) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

The following defined symbols remain to be analysed:
div

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

Proved the following rewrite lemma:
div(gen_0':s2_0(+(1, n932_0)), gen_0':s2_0(1)) → *3_0, rt ∈ Ω(n9320)

Induction Base:
div(gen_0':s2_0(+(1, 0)), gen_0':s2_0(1))

Induction Step:
div(gen_0':s2_0(+(1, +(n932_0, 1))), gen_0':s2_0(1)) →RΩ(1)
s(div(minus(gen_0':s2_0(+(1, n932_0)), gen_0':s2_0(0)), s(gen_0':s2_0(0)))) →RΩ(1)
s(div(gen_0':s2_0(+(1, n932_0)), s(gen_0':s2_0(0)))) →IH
s(*3_0)

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

(17) Complex Obligation (BEST)

(18) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)
div(gen_0':s2_0(+(1, n932_0)), gen_0':s2_0(1)) → *3_0, rt ∈ Ω(n9320)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

No more defined symbols left to analyse.

(19) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

(20) BOUNDS(n^1, INF)

(21) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)
div(gen_0':s2_0(+(1, n932_0)), gen_0':s2_0(1)) → *3_0, rt ∈ Ω(n9320)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

No more defined symbols left to analyse.

(22) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

(23) BOUNDS(n^1, INF)

(24) Obligation:

TRS:
Rules:
minus(x, 0') → x
minus(0', y) → 0'
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0') → s(s(0'))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0', y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0'))))

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
p :: 0':s → 0':s
plus :: 0':s → 0':s → 0':s
div :: 0':s → 0':s → 0':s
hole_0':s1_0 :: 0':s
gen_0':s2_0 :: Nat → 0':s

Lemmas:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

Generator Equations:
gen_0':s2_0(0) ⇔ 0'
gen_0':s2_0(+(x, 1)) ⇔ s(gen_0':s2_0(x))

No more defined symbols left to analyse.

(25) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
p(gen_0':s2_0(+(2, n4_0))) → *3_0, rt ∈ Ω(n40)

(26) BOUNDS(n^1, INF)