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:
  { terms(N) -> cons(recip(sqr(N)))
  , sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { terms(N) -> cons(recip(sqr(N))) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [terms](x1) = [3] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [0]         
                                           
        [recip](x1) = [1] x1 + [1]         
                                           
          [sqr](x1) = [2] x1 + [0]         
                                           
                [0] = [0]                  
                                           
                [s] = [0]                  
                                           
          [dbl](x1) = [2] x1 + [0]         
                                           
      [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
    [first](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
              [nil] = [0]                  
  
  This order satisfies the following ordering constraints:
  
               [terms(N)] =  [3] N + [3]          
                          >  [2] N + [1]          
                          =  [cons(recip(sqr(N)))]
                                                  
               [sqr(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [sqr(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
               [dbl(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [dbl(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
            [add(0(), X)] =  [3] X + [0]          
                          >= [1] X + [0]          
                          =  [X]                  
                                                  
            [add(s(), Y)] =  [3] Y + [0]          
                          >= [0]                  
                          =  [s()]                
                                                  
          [first(0(), X)] =  [3] X + [0]          
                          >= [0]                  
                          =  [nil()]              
                                                  
    [first(s(), cons(Y))] =  [3] Y + [0]          
                          >= [1] Y + [0]          
                          =  [cons(Y)]            
                                                  

We return to the main proof.

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

Strict Trs:
  { sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Weak Trs: { terms(N) -> cons(recip(sqr(N))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { first(s(), cons(Y)) -> cons(Y) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [terms](x1) = [3] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [2]         
                                           
        [recip](x1) = [1] x1 + [0]         
                                           
          [sqr](x1) = [2] x1 + [0]         
                                           
                [0] = [0]                  
                                           
                [s] = [0]                  
                                           
          [dbl](x1) = [2] x1 + [0]         
                                           
      [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
    [first](x1, x2) = [2] x1 + [2] x2 + [0]
                                           
              [nil] = [0]                  
  
  This order satisfies the following ordering constraints:
  
               [terms(N)] =  [3] N + [3]          
                          >  [2] N + [2]          
                          =  [cons(recip(sqr(N)))]
                                                  
               [sqr(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [sqr(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
               [dbl(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [dbl(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
            [add(0(), X)] =  [3] X + [0]          
                          >= [1] X + [0]          
                          =  [X]                  
                                                  
            [add(s(), Y)] =  [3] Y + [0]          
                          >= [0]                  
                          =  [s()]                
                                                  
          [first(0(), X)] =  [2] X + [0]          
                          >= [0]                  
                          =  [nil()]              
                                                  
    [first(s(), cons(Y))] =  [2] Y + [4]          
                          >  [1] Y + [2]          
                          =  [cons(Y)]            
                                                  

We return to the main proof.

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

Strict Trs:
  { sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , first(0(), X) -> nil() }
Weak Trs:
  { terms(N) -> cons(recip(sqr(N)))
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { first(0(), X) -> nil() }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [terms](x1) = [3] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [0]         
                                           
        [recip](x1) = [1] x1 + [2]         
                                           
          [sqr](x1) = [2] x1 + [0]         
                                           
                [0] = [0]                  
                                           
                [s] = [0]                  
                                           
          [dbl](x1) = [2] x1 + [0]         
                                           
      [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
    [first](x1, x2) = [2] x1 + [3] x2 + [1]
                                           
              [nil] = [0]                  
  
  This order satisfies the following ordering constraints:
  
               [terms(N)] =  [3] N + [3]          
                          >  [2] N + [2]          
                          =  [cons(recip(sqr(N)))]
                                                  
               [sqr(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [sqr(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
               [dbl(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [dbl(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
            [add(0(), X)] =  [3] X + [0]          
                          >= [1] X + [0]          
                          =  [X]                  
                                                  
            [add(s(), Y)] =  [3] Y + [0]          
                          >= [0]                  
                          =  [s()]                
                                                  
          [first(0(), X)] =  [3] X + [1]          
                          >  [0]                  
                          =  [nil()]              
                                                  
    [first(s(), cons(Y))] =  [3] Y + [1]          
                          >  [1] Y + [0]          
                          =  [cons(Y)]            
                                                  

We return to the main proof.

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

Strict Trs:
  { sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s() }
Weak Trs:
  { terms(N) -> cons(recip(sqr(N)))
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { sqr(s()) -> s()
  , dbl(s()) -> s()
  , add(s(), Y) -> s() }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [terms](x1) = [3] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [3]         
                                           
        [recip](x1) = [1] x1 + [0]         
                                           
          [sqr](x1) = [2] x1 + [0]         
                                           
                [0] = [0]                  
                                           
                [s] = [2]                  
                                           
          [dbl](x1) = [2] x1 + [0]         
                                           
      [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
    [first](x1, x2) = [1] x1 + [1] x2 + [2]
                                           
              [nil] = [1]                  
  
  This order satisfies the following ordering constraints:
  
               [terms(N)] =  [3] N + [3]          
                          >= [2] N + [3]          
                          =  [cons(recip(sqr(N)))]
                                                  
               [sqr(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [sqr(s())] =  [4]                  
                          >  [2]                  
                          =  [s()]                
                                                  
               [dbl(0())] =  [0]                  
                          >= [0]                  
                          =  [0()]                
                                                  
               [dbl(s())] =  [4]                  
                          >  [2]                  
                          =  [s()]                
                                                  
            [add(0(), X)] =  [3] X + [0]          
                          >= [1] X + [0]          
                          =  [X]                  
                                                  
            [add(s(), Y)] =  [3] Y + [4]          
                          >  [2]                  
                          =  [s()]                
                                                  
          [first(0(), X)] =  [1] X + [2]          
                          >  [1]                  
                          =  [nil()]              
                                                  
    [first(s(), cons(Y))] =  [1] Y + [7]          
                          >  [1] Y + [3]          
                          =  [cons(Y)]            
                                                  

We return to the main proof.

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

Strict Trs:
  { sqr(0()) -> 0()
  , dbl(0()) -> 0()
  , add(0(), X) -> X }
Weak Trs:
  { terms(N) -> cons(recip(sqr(N)))
  , sqr(s()) -> s()
  , dbl(s()) -> s()
  , add(s(), Y) -> s()
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { sqr(0()) -> 0()
  , dbl(0()) -> 0()
  , add(0(), X) -> X }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [terms](x1) = [3] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [3]         
                                           
        [recip](x1) = [1] x1 + [0]         
                                           
          [sqr](x1) = [2] x1 + [0]         
                                           
                [0] = [2]                  
                                           
                [s] = [0]                  
                                           
          [dbl](x1) = [2] x1 + [0]         
                                           
      [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                           
    [first](x1, x2) = [3] x1 + [1] x2 + [1]
                                           
              [nil] = [3]                  
  
  This order satisfies the following ordering constraints:
  
               [terms(N)] =  [3] N + [3]          
                          >= [2] N + [3]          
                          =  [cons(recip(sqr(N)))]
                                                  
               [sqr(0())] =  [4]                  
                          >  [2]                  
                          =  [0()]                
                                                  
               [sqr(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
               [dbl(0())] =  [4]                  
                          >  [2]                  
                          =  [0()]                
                                                  
               [dbl(s())] =  [0]                  
                          >= [0]                  
                          =  [s()]                
                                                  
            [add(0(), X)] =  [3] X + [4]          
                          >  [1] X + [0]          
                          =  [X]                  
                                                  
            [add(s(), Y)] =  [3] Y + [0]          
                          >= [0]                  
                          =  [s()]                
                                                  
          [first(0(), X)] =  [1] X + [7]          
                          >  [3]                  
                          =  [nil()]              
                                                  
    [first(s(), cons(Y))] =  [1] Y + [4]          
                          >  [1] Y + [3]          
                          =  [cons(Y)]            
                                                  

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:
  { terms(N) -> cons(recip(sqr(N)))
  , sqr(0()) -> 0()
  , sqr(s()) -> s()
  , dbl(0()) -> 0()
  , dbl(s()) -> s()
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , first(0(), X) -> nil()
  , first(s(), cons(Y)) -> cons(Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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