(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(int(s(x), y))
if(false, x, y) → nil
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)