The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

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

*(@x, @y) → #mult(@x, @y)
dyade(@l1, @l2) → dyade#1(@l1, @l2)
dyade#1(::(@x, @xs), @l2) → ::(mult(@x, @l2), dyade(@xs, @l2))
dyade#1(nil, @l2) → nil
mult(@n, @l) → mult#1(@l, @n)
mult#1(::(@x, @xs), @n) → ::(*(@n, @x), mult(@n, @xs))
mult#1(nil, @n) → nil

The (relative) TRS S consists of the following rules:

#add(#0, @y) → @y
#add(#neg(#s(#0)), @y) → #pred(@y)
#add(#neg(#s(#s(@x))), @y) → #pred(#add(#pos(#s(@x)), @y))
#add(#pos(#s(#0)), @y) → #succ(@y)
#add(#pos(#s(#s(@x))), @y) → #succ(#add(#pos(#s(@x)), @y))
#mult(#0, #0) → #0
#mult(#0, #neg(@y)) → #0
#mult(#0, #pos(@y)) → #0
#mult(#neg(@x), #0) → #0
#mult(#neg(@x), #neg(@y)) → #pos(#natmult(@x, @y))
#mult(#neg(@x), #pos(@y)) → #neg(#natmult(@x, @y))
#mult(#pos(@x), #0) → #0
#mult(#pos(@x), #neg(@y)) → #neg(#natmult(@x, @y))
#mult(#pos(@x), #pos(@y)) → #pos(#natmult(@x, @y))
#natmult(#0, @y) → #0
#natmult(#s(@x), @y) → #add(#pos(@y), #natmult(@x, @y))
#pred(#0) → #neg(#s(#0))
#pred(#neg(#s(@x))) → #neg(#s(#s(@x)))
#pred(#pos(#s(#0))) → #0
#pred(#pos(#s(#s(@x)))) → #pos(#s(@x))
#succ(#0) → #pos(#s(#0))
#succ(#neg(#s(#0))) → #0
#succ(#neg(#s(#s(@x)))) → #neg(#s(@x))
#succ(#pos(#s(@x))) → #pos(#s(#s(@x)))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
dyade(::(@x67605_3, @xs67606_3), @l2) →⁺ ::(mult(@x67605_3, @l2), dyade(@xs67606_3, @l2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [@xs67606_3 / ::(@x67605_3, @xs67606_3)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)