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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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:

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

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(4) Obligation:

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

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

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

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

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

(6) Obligation:

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

Types:
minus :: 0':s → 0':s → 0':s
0' :: 0':s
s :: 0':s → 0':s
quot :: 0':s → 0':s → 0':s
plus :: 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:
minus, quot, plus

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

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

Proved the following rewrite lemma:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)

Induction Base:
minus(gen_0':s2_0(0), gen_0':s2_0(0)) →_R^Ω(1)
gen_0':s2_0(0)

Induction Step:
minus(gen_0':s2_0(+(n4_0, 1)), gen_0':s2_0(+(n4_0, 1))) →_R^Ω(1)
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) →_IH
gen_0':s2_0(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:
minus(x, 0') → x
minus(s(x), s(y)) → minus(x, y)
quot(0', s(y)) → 0'
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0')), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0')))
plus(plus(x, s(0')), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0')))

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

Lemmas:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + 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:
quot, plus

(10) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(11) Obligation:

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

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

Lemmas:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + 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

(12) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
plus(gen_0':s2_0(n318_0), gen_0':s2_0(b)) → gen_0':s2_0(+(n318_0, b)), rt ∈ Ω(1 + n318₀)

Induction Base:
plus(gen_0':s2_0(0), gen_0':s2_0(b)) →_R^Ω(1)
gen_0':s2_0(b)

Induction Step:
plus(gen_0':s2_0(+(n318_0, 1)), gen_0':s2_0(b)) →_R^Ω(1)
s(plus(gen_0':s2_0(n318_0), gen_0':s2_0(b))) →_IH
s(gen_0':s2_0(+(b, c319_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:
minus(x, 0') → x
minus(s(x), s(y)) → minus(x, y)
quot(0', s(y)) → 0'
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0', y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0')), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0')))
plus(plus(x, s(0')), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0')))

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

Lemmas:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)
plus(gen_0':s2_0(n318_0), gen_0':s2_0(b)) → gen_0':s2_0(+(n318_0, b)), rt ∈ Ω(1 + n318₀)

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.

(15) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)

(16) BOUNDS(n^1, INF)

(17) Obligation:

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

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

Lemmas:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)
plus(gen_0':s2_0(n318_0), gen_0':s2_0(b)) → gen_0':s2_0(+(n318_0, b)), rt ∈ Ω(1 + n318₀)

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.

(18) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)

(19) BOUNDS(n^1, INF)

(20) Obligation:

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

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

Lemmas:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + 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.

(21) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
minus(gen_0':s2_0(n4_0), gen_0':s2_0(n4_0)) → gen_0':s2_0(0), rt ∈ Ω(1 + n4₀)

(22) BOUNDS(n^1, INF)