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:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, 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:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
./0
remove/0
if/0
=/0
=/1

(4) Obligation:

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

purge(nil) → nil
purge(.(y)) → .(purge(remove(y)))
remove(nil) → nil
remove(.(z)) → if(remove(z), .(y, remove(z)))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

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

The rewrite sequence
remove(.(z)) →⁺ if(remove(z), .(y, remove(z)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1].
The pumping substitution is [z / .(z)].
The result substitution is [ ].

(6) BOUNDS(2^n, INF)