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:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N
  , plus(N, s(M)) -> s(plus(N, M)) }
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:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N }

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](x1, x2) = [1] x1 + [3] x2 + [3]
                                          
              [tt] = [3]                  
                                          
    [activate](x1) = [3] x1 + [3]         
                                          
    [plus](x1, x2) = [3] x1 + [3] x2 + [1]
                                          
               [0] = [1]                  
                                          
           [s](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
     [and(tt(), X)] =  [3] X + [6]        
                    >  [3] X + [3]        
                    =  [activate(X)]      
                                          
      [activate(X)] =  [3] X + [3]        
                    >  [1] X + [0]        
                    =  [X]                
                                          
     [plus(N, 0())] =  [3] N + [4]        
                    >  [1] N + [0]        
                    =  [N]                
                                          
    [plus(N, s(M))] =  [3] N + [3] M + [1]
                    >= [3] N + [3] M + [1]
                    =  [s(plus(N, M))]    
                                          

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: { plus(N, s(M)) -> s(plus(N, M)) }
Weak Trs:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N }
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: { plus(N, s(M)) -> s(plus(N, M)) }

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](x1, x2) = [1] x1 + [3] x2 + [3]
                                          
              [tt] = [3]                  
                                          
    [activate](x1) = [3] x1 + [1]         
                                          
    [plus](x1, x2) = [3] x1 + [2] x2 + [0]
                                          
               [0] = [3]                  
                                          
           [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
     [and(tt(), X)] = [3] X + [6]        
                    > [3] X + [1]        
                    = [activate(X)]      
                                         
      [activate(X)] = [3] X + [1]        
                    > [1] X + [0]        
                    = [X]                
                                         
     [plus(N, 0())] = [3] N + [6]        
                    > [1] N + [0]        
                    = [N]                
                                         
    [plus(N, s(M))] = [3] N + [2] M + [4]
                    > [3] N + [2] M + [2]
                    = [s(plus(N, M))]    
                                         

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:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N
  , plus(N, s(M)) -> s(plus(N, M)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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