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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
lt(s(x), s(y)) →⁺ lt(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 [ ].

(4) BOUNDS(n^1, INF)