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:
  { f(0()) -> cons(0())
  , f(s(0())) -> f(p(s(0())))
  , p(s(X)) -> X }
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: { p(s(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).
  
       [f](x1) = [2] x1 + [0]
                             
           [0] = [0]         
                             
    [cons](x1) = [1] x1 + [0]
                             
       [s](x1) = [1] x1 + [2]
                             
       [p](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [f(0())] =  [0]           
                >= [0]           
                =  [cons(0())]   
                                 
    [f(s(0()))] =  [4]           
                >= [4]           
                =  [f(p(s(0())))]
                                 
      [p(s(X))] =  [1] X + [2]   
                >  [1] X + [0]   
                =  [X]           
                                 

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:
  { f(0()) -> cons(0())
  , f(s(0())) -> f(p(s(0()))) }
Weak Trs: { p(s(X)) -> X }
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: { f(0()) -> cons(0()) }

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).
  
       [f](x1) = [2] x1 + [1]
                             
           [0] = [0]         
                             
    [cons](x1) = [1] x1 + [0]
                             
       [s](x1) = [1] x1 + [0]
                             
       [p](x1) = [2] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [f(0())] =  [1]           
                >  [0]           
                =  [cons(0())]   
                                 
    [f(s(0()))] =  [1]           
                >= [1]           
                =  [f(p(s(0())))]
                                 
      [p(s(X))] =  [2] X + [0]   
                >= [1] X + [0]   
                =  [X]           
                                 

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: { f(s(0())) -> f(p(s(0()))) }
Weak Trs:
  { f(0()) -> cons(0())
  , p(s(X)) -> X }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(s(0())) -> f(p(s(0()))) }

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) and not(IDA(1)).
  
       [f](x1) = [2 2] x1 + [0]
                 [0 0]      [0]
                               
           [0] = [0]           
                 [0]           
                               
    [cons](x1) = [1 0] x1 + [0]
                 [0 0]      [0]
                               
       [s](x1) = [1 1] x1 + [0]
                 [0 0]      [2]
                               
       [p](x1) = [2 0] x1 + [0]
                 [1 0]      [0]
  
  This order satisfies the following ordering constraints:
  
       [f(0())] =  [0]           
                   [0]           
                >= [0]           
                   [0]           
                =  [cons(0())]   
                                 
    [f(s(0()))] =  [4]           
                   [0]           
                >  [0]           
                   [0]           
                =  [f(p(s(0())))]
                                 
      [p(s(X))] =  [2 2] X + [0] 
                   [1 1]     [0] 
                >= [1 0] X + [0] 
                   [0 1]     [0] 
                =  [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:
  { f(0()) -> cons(0())
  , f(s(0())) -> f(p(s(0())))
  , p(s(X)) -> 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))