WORST_CASE(?,O(n^1))
* Step 1: Bounds WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            group3(l) -> group3#1(l)
            group3#1(cons(x,xs)) -> group3#2(xs,x)
            group3#1(nil()) -> nil()
            group3#2(cons(y,ys),x) -> group3#3(ys,x,y)
            group3#2(nil(),x) -> nil()
            group3#3(cons(z,zs),x,y) -> cons(tuple#3(x,y,z),group3(zs))
            group3#3(nil(),x,y) -> nil()
            zip3(l1,l2,l3) -> zip3#1(l1,l2,l3)
            zip3#1(cons(x,xs),l2,l3) -> zip3#2(l2,l3,x,xs)
            zip3#1(nil(),l2,l3) -> nil()
            zip3#2(cons(y,ys),l3,x,xs) -> zip3#3(l3,x,xs,y,ys)
            zip3#2(nil(),l3,x,xs) -> nil()
            zip3#3(cons(z,zs),x,xs,y,ys) -> cons(tuple#3(x,y,z),zip3(xs,ys,zs))
            zip3#3(nil(),x,xs,y,ys) -> nil()
        - Signature:
            {group3/1,group3#1/1,group3#2/2,group3#3/3,zip3/3,zip3#1/3,zip3#2/4,zip3#3/5} / {cons/2,nil/0,tuple#3/3}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {group3,group3#1,group3#2,group3#3,zip3,zip3#1,zip3#2
            ,zip3#3} and constructors {cons,nil,tuple#3}
    + Applied Processor:
        Bounds {initialAutomaton = minimal, enrichment = match}
    + Details:
        The problem is match-bounded by 2.
        The enriched problem is compatible with follwoing automaton.
          cons_0(2,2) -> 2
          cons_1(3,4) -> 1
          cons_1(3,4) -> 4
          group3_0(2) -> 1
          group3_1(2) -> 4
          group3#1_0(2) -> 1
          group3#1_1(2) -> 1
          group3#1_2(2) -> 4
          group3#2_0(2,2) -> 1
          group3#2_1(2,2) -> 1
          group3#2_1(2,2) -> 4
          group3#3_0(2,2,2) -> 1
          group3#3_1(2,2,2) -> 1
          group3#3_1(2,2,2) -> 4
          nil_0() -> 2
          nil_1() -> 1
          nil_1() -> 4
          tuple#3_0(2,2,2) -> 2
          tuple#3_1(2,2,2) -> 3
          zip3_0(2,2,2) -> 1
          zip3_1(2,2,2) -> 4
          zip3#1_0(2,2,2) -> 1
          zip3#1_1(2,2,2) -> 1
          zip3#1_2(2,2,2) -> 4
          zip3#2_0(2,2,2,2) -> 1
          zip3#2_1(2,2,2,2) -> 1
          zip3#2_1(2,2,2,2) -> 4
          zip3#3_0(2,2,2,2,2) -> 1
          zip3#3_1(2,2,2,2,2) -> 1
          zip3#3_1(2,2,2,2,2) -> 4
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            group3(l) -> group3#1(l)
            group3#1(cons(x,xs)) -> group3#2(xs,x)
            group3#1(nil()) -> nil()
            group3#2(cons(y,ys),x) -> group3#3(ys,x,y)
            group3#2(nil(),x) -> nil()
            group3#3(cons(z,zs),x,y) -> cons(tuple#3(x,y,z),group3(zs))
            group3#3(nil(),x,y) -> nil()
            zip3(l1,l2,l3) -> zip3#1(l1,l2,l3)
            zip3#1(cons(x,xs),l2,l3) -> zip3#2(l2,l3,x,xs)
            zip3#1(nil(),l2,l3) -> nil()
            zip3#2(cons(y,ys),l3,x,xs) -> zip3#3(l3,x,xs,y,ys)
            zip3#2(nil(),l3,x,xs) -> nil()
            zip3#3(cons(z,zs),x,xs,y,ys) -> cons(tuple#3(x,y,z),zip3(xs,ys,zs))
            zip3#3(nil(),x,xs,y,ys) -> nil()
        - Signature:
            {group3/1,group3#1/1,group3#2/2,group3#3/3,zip3/3,zip3#1/3,zip3#2/4,zip3#3/5} / {cons/2,nil/0,tuple#3/3}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {group3,group3#1,group3#2,group3#3,zip3,zip3#1,zip3#2
            ,zip3#3} and constructors {cons,nil,tuple#3}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))