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:

f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(0, 1, s(z8_2)) →⁺ s(f(0, 1, z8_2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [z8_2 / s(z8_2)].
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:

f(0', 1', x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0', 1', z))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
f(0', 1', x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0', 1', z))

Types:
f :: 0':1':s → 0':1':s → 0':1':s → 0':1':s
0' :: 0':1':s
1' :: 0':1':s
s :: 0':1':s → 0':1':s
hole_0':1':s1_0 :: 0':1':s
gen_0':1':s2_0 :: Nat → 0':1':s

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

Heuristically decided to analyse the following defined symbols:
f

(8) Obligation:

TRS:
Rules:
f(0', 1', x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0', 1', z))

Types:
f :: 0':1':s → 0':1':s → 0':1':s → 0':1':s
0' :: 0':1':s
1' :: 0':1':s
s :: 0':1':s → 0':1':s
hole_0':1':s1_0 :: 0':1':s
gen_0':1':s2_0 :: Nat → 0':1':s

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

The following defined symbols remain to be analysed:
f

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

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

(10) Obligation:

TRS:
Rules:
f(0', 1', x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0', 1', z))

Types:
f :: 0':1':s → 0':1':s → 0':1':s → 0':1':s
0' :: 0':1':s
1' :: 0':1':s
s :: 0':1':s → 0':1':s
hole_0':1':s1_0 :: 0':1':s
gen_0':1':s2_0 :: Nat → 0':1':s

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

No more defined symbols left to analyse.