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:

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

+'(+'(x, y), z) → +'(x, +'(y, z))
+'(f(x), f(y)) → f(+'(x, y))
+'(f(x), +'(f(y), z)) → +'(f(+'(x, y)), z)

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
+'(+'(x, y), z) → +'(x, +'(y, z))
+'(f(x), f(y)) → f(+'(x, y))
+'(f(x), +'(f(y), z)) → +'(f(+'(x, y)), z)

Types:
+' :: f → f → f
f :: f → f
hole_f1_0 :: f
gen_f2_0 :: Nat → f

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

Heuristically decided to analyse the following defined symbols:
+'

(8) Obligation:

TRS:
Rules:
+'(+'(x, y), z) → +'(x, +'(y, z))
+'(f(x), f(y)) → f(+'(x, y))
+'(f(x), +'(f(y), z)) → +'(f(+'(x, y)), z)

Types:
+' :: f → f → f
f :: f → f
hole_f1_0 :: f
gen_f2_0 :: Nat → f

Generator Equations:
gen_f2_0(0) ⇔ hole_f1_0
gen_f2_0(+(x, 1)) ⇔ f(gen_f2_0(x))

The following defined symbols remain to be analysed:
+'

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

Could not prove a rewrite lemma for the defined symbol +'.

(10) Obligation:

TRS:
Rules:
+'(+'(x, y), z) → +'(x, +'(y, z))
+'(f(x), f(y)) → f(+'(x, y))
+'(f(x), +'(f(y), z)) → +'(f(+'(x, y)), z)

Types:
+' :: f → f → f
f :: f → f
hole_f1_0 :: f
gen_f2_0 :: Nat → f

Generator Equations:
gen_f2_0(0) ⇔ hole_f1_0
gen_f2_0(+(x, 1)) ⇔ f(gen_f2_0(x))

No more defined symbols left to analyse.