YES(O(1),O(n^1))

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

Strict Trs:
  { duplicate(Cons(x, xs)) -> Cons(x, Cons(x, duplicate(xs)))
  , duplicate(Nil()) -> Nil()
  , goal(x) -> duplicate(x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(Cons) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [duplicate](x1) = [1] x1 + [3]
                                
   [Cons](x1, x2) = [1] x2 + [3]
                                
            [Nil] = [3]         
                                
       [goal](x1) = [1] x1 + [7]

The following symbols are considered usable

  {duplicate, goal}

The order satisfies the following ordering constraints:

  [duplicate(Cons(x, xs))] = [1] xs + [6]                     
                           ? [1] xs + [9]                     
                           = [Cons(x, Cons(x, duplicate(xs)))]
                                                              
        [duplicate(Nil())] = [6]                              
                           > [3]                              
                           = [Nil()]                          
                                                              
                 [goal(x)] = [1] x + [7]                      
                           > [1] x + [3]                      
                           = [duplicate(x)]                   
                                                              

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict Trs:
  { duplicate(Cons(x, xs)) -> Cons(x, Cons(x, duplicate(xs))) }
Weak Trs:
  { duplicate(Nil()) -> Nil()
  , goal(x) -> duplicate(x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs: { duplicate(Cons(x, xs)) -> Cons(x, Cons(x, duplicate(xs))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(duplicate) = {}, safe(Cons) = {1, 2}, safe(Nil) = {},
   safe(goal) = {}
  
  and precedence
  
   duplicate ~ goal .
  
  Following symbols are considered recursive:
  
   {duplicate}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
    duplicate(Cons(; x,  xs);) >  Cons(; x,  Cons(; x,  duplicate(xs;)))
                                                                        
             duplicate(Nil();) >  Nil()                                 
                                                                        
                      goal(x;) >= duplicate(x;)                         
                                                                        

We return to the main proof.

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

Weak Trs:
  { duplicate(Cons(x, xs)) -> Cons(x, Cons(x, duplicate(xs)))
  , duplicate(Nil()) -> Nil()
  , goal(x) -> duplicate(x) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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