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(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, 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(g(x), y, z) →⁺ g(f(x, y, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / g(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:

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

Types:
f :: 0':1':g → 0':1':g → 0':1':g → 0':1':g
0' :: 0':1':g
1' :: 0':1':g
g :: 0':1':g → 0':1':g
hole_0':1':g1_0 :: 0':1':g
gen_0':1':g2_0 :: Nat → 0':1':g

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

Heuristically decided to analyse the following defined symbols:
f

(8) Obligation:

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

Types:
f :: 0':1':g → 0':1':g → 0':1':g → 0':1':g
0' :: 0':1':g
1' :: 0':1':g
g :: 0':1':g → 0':1':g
hole_0':1':g1_0 :: 0':1':g
gen_0':1':g2_0 :: Nat → 0':1':g

Generator Equations:
gen_0':1':g2_0(0) ⇔ 1'
gen_0':1':g2_0(+(x, 1)) ⇔ g(gen_0':1':g2_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(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Types:
f :: 0':1':g → 0':1':g → 0':1':g → 0':1':g
0' :: 0':1':g
1' :: 0':1':g
g :: 0':1':g → 0':1':g
hole_0':1':g1_0 :: 0':1':g
gen_0':1':g2_0 :: Nat → 0':1':g

Generator Equations:
gen_0':1':g2_0(0) ⇔ 1'
gen_0':1':g2_0(+(x, 1)) ⇔ g(gen_0':1':g2_0(x))

No more defined symbols left to analyse.