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:
  { 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [half](x1) = [1] x1 + [0]
                             
           [0] = [0]         
                             
       [s](x1) = [1] x1 + [0]
                             
     [log](x1) = [2] x1 + [1]
  
  This order satisfies the following ordering constraints:
  
        [half(0())] =  [0]                 
                    >= [0]                 
                    =  [0()]               
                                           
    [half(s(s(x)))] =  [1] x + [0]         
                    >= [1] x + [0]         
                    =  [s(half(x))]        
                                           
      [log(s(0()))] =  [1]                 
                    >  [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [2] x + [1]         
                    >= [2] x + [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)) }

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).
  
    [half](x1) = [1] x1 + [0]
                             
           [0] = [0]         
                             
       [s](x1) = [1] x1 + [1]
                             
     [log](x1) = [1] x1 + [3]
  
  This 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()))] =  [4]                 
                    >  [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [1] x + [5]         
                    >= [1] x + [5]         
                    =  [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()
  , log(s(s(x))) -> s(log(s(half(x)))) }
Weak Trs:
  { half(s(s(x))) -> s(half(x))
  , 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: { 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:
----------
  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) = [2] x1 + [0]
  
  This 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()))] =  [2]                 
                    >  [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [2] x + [4]         
                    >  [2] x + [3]         
                    =  [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 2' 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [half](x1) = [1 0] x1 + [1]
                 [0 1]      [0]
                               
           [0] = [0]           
                 [0]           
                               
       [s](x1) = [1 2] x1 + [0]
                 [0 0]      [1]
                               
     [log](x1) = [2 0] x1 + [0]
                 [0 0]      [1]
  
  This order satisfies the following ordering constraints:
  
        [half(0())] =  [1]                 
                       [0]                 
                    >  [0]                 
                       [0]                 
                    =  [0()]               
                                           
    [half(s(s(x)))] =  [1 2] x + [3]       
                       [0 0]     [1]       
                    >  [1 2] x + [1]       
                       [0 0]     [1]       
                    =  [s(half(x))]        
                                           
      [log(s(0()))] =  [0]                 
                       [1]                 
                    >= [0]                 
                       [0]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [2 4] x + [4]       
                       [0 0]     [1]       
                    >= [2 4] x + [4]       
                       [0 0]     [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(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))