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:
  { not(true()) -> false()
  , not(false()) -> true()
  , evenodd(x, 0()) -> not(evenodd(x, s(0())))
  , evenodd(0(), s(0())) -> false()
  , evenodd(s(x), s(0())) -> evenodd(x, 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: { evenodd(0(), s(0())) -> false() }

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).
  
            [not](x1) = [1] x1 + [0]         
                                             
               [true] = [0]                  
                                             
              [false] = [0]                  
                                             
    [evenodd](x1, x2) = [2] x1 + [1] x2 + [1]
                                             
                  [0] = [0]                  
                                             
              [s](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
              [not(true())] =  [0]                      
                            >= [0]                      
                            =  [false()]                
                                                        
             [not(false())] =  [0]                      
                            >= [0]                      
                            =  [true()]                 
                                                        
          [evenodd(x, 0())] =  [2] x + [1]              
                            >= [2] x + [1]              
                            =  [not(evenodd(x, s(0())))]
                                                        
     [evenodd(0(), s(0()))] =  [1]                      
                            >  [0]                      
                            =  [false()]                
                                                        
    [evenodd(s(x), s(0()))] =  [2] x + [1]              
                            >= [2] x + [1]              
                            =  [evenodd(x, 0())]        
                                                        

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:
  { not(true()) -> false()
  , not(false()) -> true()
  , evenodd(x, 0()) -> not(evenodd(x, s(0())))
  , evenodd(s(x), s(0())) -> evenodd(x, 0()) }
Weak Trs: { evenodd(0(), s(0())) -> false() }
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: { evenodd(s(x), s(0())) -> evenodd(x, 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)).
  
            [not](x1) = [1 0] x1 + [0]           
                        [0 0]      [0]           
                                                 
               [true] = [0]                      
                        [0]                      
                                                 
              [false] = [0]                      
                        [0]                      
                                                 
    [evenodd](x1, x2) = [2 2] x1 + [2 0] x2 + [0]
                        [0 0]      [0 0]      [0]
                                                 
                  [0] = [0]                      
                        [0]                      
                                                 
              [s](x1) = [1 2] x1 + [0]           
                        [0 0]      [2]           
  
  This order satisfies the following ordering constraints:
  
              [not(true())] =  [0]                      
                               [0]                      
                            >= [0]                      
                               [0]                      
                            =  [false()]                
                                                        
             [not(false())] =  [0]                      
                               [0]                      
                            >= [0]                      
                               [0]                      
                            =  [true()]                 
                                                        
          [evenodd(x, 0())] =  [2 2] x + [0]            
                               [0 0]     [0]            
                            >= [2 2] x + [0]            
                               [0 0]     [0]            
                            =  [not(evenodd(x, s(0())))]
                                                        
     [evenodd(0(), s(0()))] =  [0]                      
                               [0]                      
                            >= [0]                      
                               [0]                      
                            =  [false()]                
                                                        
    [evenodd(s(x), s(0()))] =  [2 4] x + [4]            
                               [0 0]     [0]            
                            >  [2 2] x + [0]            
                               [0 0]     [0]            
                            =  [evenodd(x, 0())]        
                                                        

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:
  { not(true()) -> false()
  , not(false()) -> true()
  , evenodd(x, 0()) -> not(evenodd(x, s(0()))) }
Weak Trs:
  { evenodd(0(), s(0())) -> false()
  , evenodd(s(x), s(0())) -> evenodd(x, 0()) }
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: { evenodd(x, 0()) -> not(evenodd(x, 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)).
  
            [not](x1) = [1 0] x1 + [0]           
                        [0 0]      [0]           
                                                 
               [true] = [0]                      
                        [0]                      
                                                 
              [false] = [0]                      
                        [0]                      
                                                 
    [evenodd](x1, x2) = [2 0] x1 + [2 3] x2 + [2]
                        [0 0]      [0 0]      [0]
                                                 
                  [0] = [0]                      
                        [1]                      
                                                 
              [s](x1) = [1 0] x1 + [1]           
                        [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
              [not(true())] =  [0]                      
                               [0]                      
                            >= [0]                      
                               [0]                      
                            =  [false()]                
                                                        
             [not(false())] =  [0]                      
                               [0]                      
                            >= [0]                      
                               [0]                      
                            =  [true()]                 
                                                        
          [evenodd(x, 0())] =  [2 0] x + [5]            
                               [0 0]     [0]            
                            >  [2 0] x + [4]            
                               [0 0]     [0]            
                            =  [not(evenodd(x, s(0())))]
                                                        
     [evenodd(0(), s(0()))] =  [4]                      
                               [0]                      
                            >  [0]                      
                               [0]                      
                            =  [false()]                
                                                        
    [evenodd(s(x), s(0()))] =  [2 0] x + [6]            
                               [0 0]     [0]            
                            >  [2 0] x + [5]            
                               [0 0]     [0]            
                            =  [evenodd(x, 0())]        
                                                        

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:
  { not(true()) -> false()
  , not(false()) -> true() }
Weak Trs:
  { evenodd(x, 0()) -> not(evenodd(x, s(0())))
  , evenodd(0(), s(0())) -> false()
  , evenodd(s(x), s(0())) -> evenodd(x, 0()) }
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:
  { not(true()) -> false()
  , not(false()) -> true() }

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)).
  
            [not](x1) = [1 0] x1 + [1]           
                        [0 0]      [0]           
                                                 
               [true] = [0]                      
                        [0]                      
                                                 
              [false] = [0]                      
                        [0]                      
                                                 
    [evenodd](x1, x2) = [2 0] x1 + [1 1] x2 + [0]
                        [0 0]      [0 0]      [0]
                                                 
                  [0] = [1]                      
                        [3]                      
                                                 
              [s](x1) = [1 0] x1 + [1]           
                        [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
              [not(true())] =  [1]                      
                               [0]                      
                            >  [0]                      
                               [0]                      
                            =  [false()]                
                                                        
             [not(false())] =  [1]                      
                               [0]                      
                            >  [0]                      
                               [0]                      
                            =  [true()]                 
                                                        
          [evenodd(x, 0())] =  [2 0] x + [4]            
                               [0 0]     [0]            
                            >  [2 0] x + [3]            
                               [0 0]     [0]            
                            =  [not(evenodd(x, s(0())))]
                                                        
     [evenodd(0(), s(0()))] =  [4]                      
                               [0]                      
                            >  [0]                      
                               [0]                      
                            =  [false()]                
                                                        
    [evenodd(s(x), s(0()))] =  [2 0] x + [4]            
                               [0 0]     [0]            
                            >= [2 0] x + [4]            
                               [0 0]     [0]            
                            =  [evenodd(x, 0())]        
                                                        

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:
  { not(true()) -> false()
  , not(false()) -> true()
  , evenodd(x, 0()) -> not(evenodd(x, s(0())))
  , evenodd(0(), s(0())) -> false()
  , evenodd(s(x), s(0())) -> evenodd(x, 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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