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:

f(0) → 1
f(s(x)) → g(f(x))
g(x) → +(x, s(x))
f(s(x)) → +(f(x), s(f(x)))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
f(s(s(x6_2))) →⁺ g(+(f(x6_2), s(f(x6_2))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [x6_2 / s(s(x6_2))].
The result substitution is [ ].

The rewrite sequence
f(s(s(x6_2))) →⁺ g(+(f(x6_2), s(f(x6_2))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
The pumping substitution is [x6_2 / s(s(x6_2))].
The result substitution is [ ].

(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:

f(0') → 1'
f(s(x)) → g(f(x))
g(x) → +'(x, s(x))
f(s(x)) → +'(f(x), s(f(x)))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
f(0') → 1'
f(s(x)) → g(f(x))
g(x) → +'(x, s(x))
f(s(x)) → +'(f(x), s(f(x)))

Types:
f :: 0':1':s:+' → 0':1':s:+'
0' :: 0':1':s:+'
1' :: 0':1':s:+'
s :: 0':1':s:+' → 0':1':s:+'
g :: 0':1':s:+' → 0':1':s:+'
+' :: 0':1':s:+' → 0':1':s:+' → 0':1':s:+'
hole_0':1':s:+'1_0 :: 0':1':s:+'
gen_0':1':s:+'2_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'
f(s(x)) → g(f(x))
g(x) → +'(x, s(x))
f(s(x)) → +'(f(x), s(f(x)))

Types:
f :: 0':1':s:+' → 0':1':s:+'
0' :: 0':1':s:+'
1' :: 0':1':s:+'
s :: 0':1':s:+' → 0':1':s:+'
g :: 0':1':s:+' → 0':1':s:+'
+' :: 0':1':s:+' → 0':1':s:+' → 0':1':s:+'
hole_0':1':s:+'1_0 :: 0':1':s:+'
gen_0':1':s:+'2_0 :: Nat → 0':1':s:+'

Generator Equations:
gen_0':1':s:+'2_0(0) ⇔ 0'
gen_0':1':s:+'2_0(+(x, 1)) ⇔ s(gen_0':1':s:+'2_0(x))

The following defined symbols remain to be analysed:
f

(9) RewriteLemmaProof (EQUIVALENT transformation)

Proved the following rewrite lemma:
f(gen_0':1':s:+'2_0(+(1, n4_0))) → *3_0, rt ∈ Ω(2ⁿ)

Induction Base:
f(gen_0':1':s:+'2_0(+(1, 0)))

Induction Step:
f(gen_0':1':s:+'2_0(+(1, +(n4_0, 1)))) →_R^Ω(1)
+'(f(gen_0':1':s:+'2_0(+(1, n4_0))), s(f(gen_0':1':s:+'2_0(+(1, n4_0))))) →_IH
+'(*3_0, s(f(gen_0':1':s:+'2_0(+(1, n4_0))))) →_IH
+'(*3_0, s(*3_0))

We have rt ∈ Ω(2ⁿ) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(2ⁿ)

(10) BOUNDS(2^n, INF)