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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
./0

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(y), z) → .(++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

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

(6) BOUNDS(n^1, INF)