*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        group3(@l) -> group3#1(@l)
        group3#1(::(@x,@xs)) -> group3#2(@xs,@x)
        group3#1(nil()) -> nil()
        group3#2(::(@y,@ys),@x) -> group3#3(@ys,@x,@y)
        group3#2(nil(),@x) -> nil()
        group3#3(::(@z,@zs),@x,@y) -> ::(tuple#3(@x,@y,@z),group3(@zs))
        group3#3(nil(),@x,@y) -> nil()
        zip3(@l1,@l2,@l3) -> zip3#1(@l1,@l2,@l3)
        zip3#1(::(@x,@xs),@l2,@l3) -> zip3#2(@l2,@l3,@x,@xs)
        zip3#1(nil(),@l2,@l3) -> nil()
        zip3#2(::(@y,@ys),@l3,@x,@xs) -> zip3#3(@l3,@x,@xs,@y,@ys)
        zip3#2(nil(),@l3,@x,@xs) -> nil()
        zip3#3(::(@z,@zs),@x,@xs,@y,@ys) -> ::(tuple#3(@x,@y,@z),zip3(@xs,@ys,@zs))
        zip3#3(nil(),@x,@xs,@y,@ys) -> nil()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {group3/1,group3#1/1,group3#2/2,group3#3/3,zip3/3,zip3#1/3,zip3#2/4,zip3#3/5} / {::/2,nil/0,tuple#3/3}
      Obligation:
        Innermost
        basic terms: {group3,group3#1,group3#2,group3#3,zip3,zip3#1,zip3#2,zip3#3}/{::,nil,tuple#3}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 2.
      The enriched problem is compatible with follwoing automaton.
        ::_0(2,2) -> 2
        ::_1(3,4) -> 1
        ::_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
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        group3(@l) -> group3#1(@l)
        group3#1(::(@x,@xs)) -> group3#2(@xs,@x)
        group3#1(nil()) -> nil()
        group3#2(::(@y,@ys),@x) -> group3#3(@ys,@x,@y)
        group3#2(nil(),@x) -> nil()
        group3#3(::(@z,@zs),@x,@y) -> ::(tuple#3(@x,@y,@z),group3(@zs))
        group3#3(nil(),@x,@y) -> nil()
        zip3(@l1,@l2,@l3) -> zip3#1(@l1,@l2,@l3)
        zip3#1(::(@x,@xs),@l2,@l3) -> zip3#2(@l2,@l3,@x,@xs)
        zip3#1(nil(),@l2,@l3) -> nil()
        zip3#2(::(@y,@ys),@l3,@x,@xs) -> zip3#3(@l3,@x,@xs,@y,@ys)
        zip3#2(nil(),@l3,@x,@xs) -> nil()
        zip3#3(::(@z,@zs),@x,@xs,@y,@ys) -> ::(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} / {::/2,nil/0,tuple#3/3}
      Obligation:
        Innermost
        basic terms: {group3,group3#1,group3#2,group3#3,zip3,zip3#1,zip3#2,zip3#3}/{::,nil,tuple#3}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).