(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, 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:
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))
S is empty.
Rewrite Strategy: FULL
(3) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(2n):
The rewrite sequence
rev2(x, ++(y, z)) →+ ++(rev1(x, rev(rev2(y, z))), rev2(x, rev(rev2(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
The pumping substitution is [z / ++(y, z)].
The result substitution is [x / y].
The rewrite sequence
rev2(x, ++(y, z)) →+ ++(rev1(x, rev(rev2(y, z))), rev2(x, rev(rev2(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1,0].
The pumping substitution is [z / ++(y, z)].
The result substitution is [x / y].
(4) BOUNDS(2^n, INF)