* Step 1: Sum WORST_CASE(Omega(n^1),?)
    + Considered Problem:
        - Strict TRS:
            check(free(x)) -> free(check(x))
            check(new(x)) -> new(check(x))
            check(old(x)) -> old(x)
            check(old(x)) -> old(check(x))
            new(free(x)) -> free(new(x))
            new(serve()) -> free(serve())
            old(free(x)) -> free(old(x))
            old(serve()) -> free(serve())
            top1(free(x),y) -> top2(x,check(new(y)))
            top1(free(x),y) -> top2(check(x),new(y))
            top1(free(x),y) -> top2(check(new(x)),y)
            top1(free(x),y) -> top2(new(x),check(y))
            top2(x,free(y)) -> top1(x,check(new(y)))
            top2(x,free(y)) -> top1(check(x),new(y))
            top2(x,free(y)) -> top1(check(new(x)),y)
            top2(x,free(y)) -> top1(new(x),check(y))
        - Signature:
            {check/1,new/1,old/1,top1/2,top2/2} / {free/1,serve/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {check,new,old,top1,top2} and constructors {free,serve}
    + Applied Processor:
        Sum {left = someStrategy, right = someStrategy}
    + Details:
        ()
* Step 2: DecreasingLoops WORST_CASE(Omega(n^1),?)
    + Considered Problem:
        - Strict TRS:
            check(free(x)) -> free(check(x))
            check(new(x)) -> new(check(x))
            check(old(x)) -> old(x)
            check(old(x)) -> old(check(x))
            new(free(x)) -> free(new(x))
            new(serve()) -> free(serve())
            old(free(x)) -> free(old(x))
            old(serve()) -> free(serve())
            top1(free(x),y) -> top2(x,check(new(y)))
            top1(free(x),y) -> top2(check(x),new(y))
            top1(free(x),y) -> top2(check(new(x)),y)
            top1(free(x),y) -> top2(new(x),check(y))
            top2(x,free(y)) -> top1(x,check(new(y)))
            top2(x,free(y)) -> top1(check(x),new(y))
            top2(x,free(y)) -> top1(check(new(x)),y)
            top2(x,free(y)) -> top1(new(x),check(y))
        - Signature:
            {check/1,new/1,old/1,top1/2,top2/2} / {free/1,serve/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {check,new,old,top1,top2} and constructors {free,serve}
    + Applied Processor:
        DecreasingLoops {bound = AnyLoop, narrow = 10}
    + Details:
        The system has following decreasing Loops:
          check(x){x -> free(x)} =
            check(free(x)) ->^+ free(check(x))
              = C[check(x) = check(x){}]

WORST_CASE(Omega(n^1),?)