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:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

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:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
admit/0
=/0
=/1
sum/0
sum/1
sum/2
carry/0
carry/1
carry/2

(4) Obligation:

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

admit(nil) → nil
admit(.(u, .(v, .(w, z)))) → cond(=, .(u, .(v, .(w, admit(z)))))
cond(true, y) → y

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
admit(.(u, .(v, .(w, z)))) →⁺ cond(=, .(u, .(v, .(w, admit(z)))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1,1,1].
The pumping substitution is [z / .(u, .(v, .(w, z)))].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)