YES(?,O(n^1))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { 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() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(::) = {2}
TcT has computed following constructor-restricted linear polynomial
interpretation.
                [group3](x1) = 1 + 2*x1              
                                                     
              [group3#1](x1) = 2*x1                  
                                                     
                [::](x1, x2) = 2 + x1 + x2           
                                                     
          [group3#2](x1, x2) = 2*x1                  
                                                     
                     [nil]() = 2                     
                                                     
      [group3#3](x1, x2, x3) = 2*x1                  
                                                     
       [tuple#3](x1, x2, x3) = 0                     
                                                     
          [zip3](x1, x2, x3) = 1 + 2*x1 + 2*x2 + 2*x3
                                                     
        [zip3#1](x1, x2, x3) = 2*x1 + 2*x2 + 2*x3    
                                                     
    [zip3#2](x1, x2, x3, x4) = 2*x1 + 2*x2 + 2*x4    
                                                     
[zip3#3](x1, x2, x3, x4, x5) = 2*x1 + 2*x3 + 2*x5    
                                                     

This order satisfies following ordering constraints

                             [group3(@l)] = 1 + 2*@l                                      
                                          > 2*@l                                          
                                          = [group3#1(@l)]                                
                                                                                          
                  [group3#1(::(@x, @xs))] = 4 + 2*@x + 2*@xs                              
                                          > 2*@xs                                         
                                          = [group3#2(@xs, @x)]                           
                                                                                          
                        [group3#1(nil())] = 4                                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          
              [group3#2(::(@y, @ys), @x)] = 4 + 2*@y + 2*@ys                              
                                          > 2*@ys                                         
                                          = [group3#3(@ys, @x, @y)]                       
                                                                                          
                    [group3#2(nil(), @x)] = 4                                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          
          [group3#3(::(@z, @zs), @x, @y)] = 4 + 2*@z + 2*@zs                              
                                          > 3 + 2*@zs                                     
                                          = [::(tuple#3(@x, @y, @z), group3(@zs))]        
                                                                                          
                [group3#3(nil(), @x, @y)] = 4                                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          
                    [zip3(@l1, @l2, @l3)] = 1 + 2*@l1 + 2*@l2 + 2*@l3                     
                                          > 2*@l1 + 2*@l2 + 2*@l3                         
                                          = [zip3#1(@l1, @l2, @l3)]                       
                                                                                          
          [zip3#1(::(@x, @xs), @l2, @l3)] = 4 + 2*@x + 2*@xs + 2*@l2 + 2*@l3              
                                          > 2*@l2 + 2*@l3 + 2*@xs                         
                                          = [zip3#2(@l2, @l3, @x, @xs)]                   
                                                                                          
                [zip3#1(nil(), @l2, @l3)] = 4 + 2*@l2 + 2*@l3                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          
      [zip3#2(::(@y, @ys), @l3, @x, @xs)] = 4 + 2*@y + 2*@ys + 2*@l3 + 2*@xs              
                                          > 2*@l3 + 2*@xs + 2*@ys                         
                                          = [zip3#3(@l3, @x, @xs, @y, @ys)]               
                                                                                          
            [zip3#2(nil(), @l3, @x, @xs)] = 4 + 2*@l3 + 2*@xs                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          
  [zip3#3(::(@z, @zs), @x, @xs, @y, @ys)] = 4 + 2*@z + 2*@zs + 2*@xs + 2*@ys              
                                          > 3 + 2*@xs + 2*@ys + 2*@zs                     
                                          = [::(tuple#3(@x, @y, @z), zip3(@xs, @ys, @zs))]
                                                                                          
        [zip3#3(nil(), @x, @xs, @y, @ys)] = 4 + 2*@xs + 2*@ys                             
                                          > 2                                             
                                          = [nil()]                                       
                                                                                          

Hurray, we answered YES(?,O(n^1))