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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, 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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
cons/0

(4) Obligation:

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

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, 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
le(s(x), s(y)) →⁺ le(x, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x), y / s(y)].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)