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))

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

The following argument positions are usable:
  Uargs(s) = {1}, Uargs(log) = {1}

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

  [half](x1) = [0]         
                           
         [0] = [4]         
                           
     [s](x1) = [1] x1 + [4]
                           
   [log](x1) = [1] x1 + [0]

The following symbols are considered usable

  {half, log}

The order satisfies the following ordering constraints:

      [half(0())] =  [0]                 
                  ?  [4]                 
                  =  [0()]               
                                         
  [half(s(s(x)))] =  [0]                 
                  ?  [4]                 
                  =  [s(half(x))]        
                                         
    [log(s(0()))] =  [8]                 
                  >  [4]                 
                  =  [0()]               
                                         
   [log(s(s(x)))] =  [1] x + [8]         
                  >= [8]                 
                  =  [s(log(s(half(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:
  { 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))

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

The following argument positions are usable:
  Uargs(s) = {1}, Uargs(log) = {1}

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

  [half](x1) = [1] x1 + [0]
                           
         [0] = [4]         
                           
     [s](x1) = [1] x1 + [4]
                           
   [log](x1) = [1] x1 + [0]

The following symbols are considered usable

  {half, log}

The order satisfies the following ordering constraints:

      [half(0())] =  [4]                 
                  >= [4]                 
                  =  [0()]               
                                         
  [half(s(s(x)))] =  [1] x + [8]         
                  >  [1] x + [4]         
                  =  [s(half(x))]        
                                         
    [log(s(0()))] =  [8]                 
                  >  [4]                 
                  =  [0()]               
                                         
   [log(s(s(x)))] =  [1] x + [8]         
                  >= [1] x + [8]         
                  =  [s(log(s(half(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:
  { 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))

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

The following argument positions are usable:
  Uargs(s) = {1}, Uargs(log) = {1}

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

  [half](x1) = [1] x1 + [4]
                           
         [0] = [4]         
                           
     [s](x1) = [1] x1 + [4]
                           
   [log](x1) = [1] x1 + [0]

The following symbols are considered usable

  {half, log}

The order satisfies the following ordering constraints:

      [half(0())] = [8]                 
                  > [4]                 
                  = [0()]               
                                        
  [half(s(s(x)))] = [1] x + [12]        
                  > [1] x + [8]         
                  = [s(half(x))]        
                                        
    [log(s(0()))] = [8]                 
                  > [4]                 
                  = [0()]               
                                        
   [log(s(s(x)))] = [1] x + [8]         
                  ? [1] x + [12]        
                  = [s(log(s(half(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: { log(s(s(x))) -> s(log(s(half(x)))) }
Weak Trs:
  { half(0()) -> 0()
  , 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:
----------
  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] = [2]         
                             
       [s](x1) = [1] x1 + [2]
                             
     [log](x1) = [2] x1 + [0]
  
  The following symbols are considered usable
  
    {half, log}
  
  The order satisfies the following ordering constraints:
  
        [half(0())] =  [2]                 
                    >= [2]                 
                    =  [0()]               
                                           
    [half(s(s(x)))] =  [1] x + [4]         
                    >  [1] x + [2]         
                    =  [s(half(x))]        
                                           
      [log(s(0()))] =  [8]                 
                    >  [2]                 
                    =  [0()]               
                                           
     [log(s(s(x)))] =  [2] x + [8]         
                    >  [2] x + [6]         
                    =  [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))