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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
p(m, s(r3_1), s(r)) →⁺ p(m, r3_1, r)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [r3_1 / s(r3_1), r / s(r)].
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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0') → p(0', n, m)
p(m, 0', 0') → m

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0') → p(0', n, m)
p(m, 0', 0') → m

Types:
p :: s:0' → s:0' → s:0' → s:0'
s :: s:0' → s:0'
0' :: s:0'
hole_s:0'1_0 :: s:0'
gen_s:0'2_0 :: Nat → s:0'

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

Heuristically decided to analyse the following defined symbols:
p

(8) Obligation:

TRS:
Rules:
p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0') → p(0', n, m)
p(m, 0', 0') → m

Types:
p :: s:0' → s:0' → s:0' → s:0'
s :: s:0' → s:0'
0' :: s:0'
hole_s:0'1_0 :: s:0'
gen_s:0'2_0 :: Nat → s:0'

Generator Equations:
gen_s:0'2_0(0) ⇔ 0'
gen_s:0'2_0(+(x, 1)) ⇔ s(gen_s:0'2_0(x))

The following defined symbols remain to be analysed:
p

(9) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

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

(10) Obligation:

TRS:
Rules:
p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0') → p(0', n, m)
p(m, 0', 0') → m

Types:
p :: s:0' → s:0' → s:0' → s:0'
s :: s:0' → s:0'
0' :: s:0'
hole_s:0'1_0 :: s:0'
gen_s:0'2_0 :: Nat → s:0'

Generator Equations:
gen_s:0'2_0(0) ⇔ 0'
gen_s:0'2_0(+(x, 1)) ⇔ s(gen_s:0'2_0(x))

No more defined symbols left to analyse.