(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
divide_ys#1(x2, x1) → Cons(take#2(x1, x2), Cons(drop#2(x1, x2), Nil))
cond_merge_ys_zs_2(True, Cons(x7, x8), Cons(x5, x6), x4, x3, x2, x1) → Cons(x4, merge#2(x3, Cons(x5, x6)))
cond_merge_ys_zs_2(False, Cons(x7, x8), Cons(x5, x6), x4, x3, x2, x1) → Cons(x2, merge#2(Cons(x7, x8), x1))
merge#2(Nil, x2) → x2
merge#2(Cons(x4, x2), Nil) → Cons(x4, x2)
merge#2(Cons(x8, x6), Cons(x4, x2)) → cond_merge_ys_zs_2(leq#2(x8, x4), Cons(x8, x6), Cons(x4, x2), x8, x6, x4, x2)
dc#1(map, divisible, mergesort_zs_3, divide, const_f, Nil) → Nil
dc#1(map, divisible, mergesort_zs_3, divide, const_f, Cons(x229, Nil)) → Cons(x229, Nil)
dc#1(map, divisible, mergesort_zs_3, divide, const_f, Cons(x51, Cons(x25, x33))) → const_f#2(Cons(x51, Cons(x25, x33)), map#2(dc(map, divisible, mergesort_zs_3, divide, const_f), divide_ys#1(Cons(x51, Cons(x25, x33)), S(halve#1(length#1(x33))))))
drop#2(0, x2) → x2
drop#2(S(0), Nil) → bot[1]
drop#2(S(x10), Cons(x56, x64)) → drop#2(x10, x64)
take#2(0, x2) → Nil
take#2(S(0), Nil) → Cons(bot[0], Nil)
take#2(S(x22), Cons(x56, x64)) → Cons(x56, take#2(x22, x64))
halve#1(0) → 0
halve#1(S(0)) → S(0)
halve#1(S(S(x14))) → S(halve#1(x14))
const_f#2(x3, Cons(x6, Cons(x4, x2))) → merge#2(x6, x4)
leq#2(0, x16) → True
leq#2(S(x20), 0) → False
leq#2(S(x4), S(x2)) → leq#2(x4, x2)
length#1(Nil) → 0
length#1(Cons(x6, x8)) → S(length#1(x8))
map#2(dc(x2, x4, x6, x8, x10), Nil) → Nil
map#2(dc(x6, x8, x10, x12, x14), Cons(x4, x2)) → Cons(dc#1(x6, x8, x10, x12, x14, x4), map#2(dc(x6, x8, x10, x12, x14), x2))
main(x113) → dc#1(map, divisible, mergesort_zs_3, divide, const_f, x113)
Rewrite Strategy: INNERMOST
(1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
cond_merge_ys_zs_2(True, Cons(x7, x8), Cons(x5, x6), x4, Cons(0, x669746_11), x2, x1) →+ Cons(x4, cond_merge_ys_zs_2(True, Cons(0, x669746_11), Cons(x5, x6), 0, x669746_11, x5, x6))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [x669746_11 / Cons(0, x669746_11)].
The result substitution is [x7 / 0, x8 / x669746_11, x4 / 0, x2 / x5, x1 / x6].
(2) BOUNDS(n^1, INF)