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

Strict Trs:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(x)))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The input is overlay and right-linear. Switching to innermost
rewriting.

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

Strict Trs:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(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: { log(s(0())) -> 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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}, Uargs(log) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [half](x1) = [0]         
                             
           [0] = [0]         
                             
       [s](x1) = [1] x1 + [0]
                             
     [log](x1) = [7] x1 + [1]
  
  The order satisfies the following ordering constraints:
  
        [half(0())] =  [0]                 
                    >= [0]                 
                    =  [0()]               
                                           
    [half(s(s(x)))] =  [0]                 
                    >= [0]                 
                    =  [s(half(x))]        
                                           
      [log(s(0()))] =  [1]                 
                    >  [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [7] x + [1]         
                    >= [1]                 
                    =  [s(log(s(half(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:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x))
  , log(s(s(x))) -> s(log(s(half(x)))) }
Weak Trs: { log(s(0())) -> 0() }
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:
  { half(s(s(x))) -> s(half(x))
  , log(s(s(x))) -> s(log(s(half(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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}, Uargs(log) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [half](x1) = [1] x1 + [0]
                             
           [0] = [0]         
                             
       [s](x1) = [1] x1 + [1]
                             
     [log](x1) = [5] x1 + [3]
  
  The order satisfies the following ordering constraints:
  
        [half(0())] =  [0]                 
                    >= [0]                 
                    =  [0()]               
                                           
    [half(s(s(x)))] =  [1] x + [2]         
                    >  [1] x + [1]         
                    =  [s(half(x))]        
                                           
      [log(s(0()))] =  [8]                 
                    >  [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [5] x + [13]        
                    >  [5] x + [9]         
                    =  [s(log(s(half(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: { half(0()) -> 0() }
Weak Trs:
  { half(s(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(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: { half(0()) -> 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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}, Uargs(log) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [half](x1) = [1] x1 + [1]
                             
           [0] = [4]         
                             
       [s](x1) = [1] x1 + [3]
                             
     [log](x1) = [2] x1 + [0]
  
  The order satisfies the following ordering constraints:
  
        [half(0())] = [5]                 
                    > [4]                 
                    = [0()]               
                                          
    [half(s(s(x)))] = [1] x + [7]         
                    > [1] x + [4]         
                    = [s(half(x))]        
                                          
      [log(s(0()))] = [14]                
                    > [4]                 
                    = [0()]               
                                          
     [log(s(s(x)))] = [2] x + [12]        
                    > [2] x + [11]        
                    = [s(log(s(half(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:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(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))