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

Strict Trs:
  { implies(x, or(y, z)) -> or(y, implies(x, z))
  , implies(not(x), y) -> or(x, y)
  , implies(not(x), or(y, z)) -> implies(y, or(x, z)) }
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:
  { implies(not(x), y) -> or(x, y)
  , implies(not(x), or(y, z)) -> implies(y, or(x, z)) }

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(or) = {2}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [implies](x1, x2) = [2] x1 + [2] x2 + [0]
                                             
            [not](x1) = [1] x1 + [4]         
                                             
         [or](x1, x2) = [1] x1 + [1] x2 + [0]
  
  The order satisfies the following ordering constraints:
  
         [implies(x, or(y, z))] =  [2] x + [2] y + [2] z + [0]
                                >= [2] x + [1] y + [2] z + [0]
                                =  [or(y, implies(x, z))]     
                                                              
           [implies(not(x), y)] =  [2] x + [2] y + [8]        
                                >  [1] x + [1] y + [0]        
                                =  [or(x, y)]                 
                                                              
    [implies(not(x), or(y, z))] =  [2] x + [2] y + [2] z + [8]
                                >  [2] x + [2] y + [2] z + [0]
                                =  [implies(y, or(x, z))]     
                                                              

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: { implies(x, or(y, z)) -> or(y, implies(x, z)) }
Weak Trs:
  { implies(not(x), y) -> or(x, y)
  , implies(not(x), or(y, z)) -> implies(y, or(x, z)) }
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: { implies(x, or(y, z)) -> or(y, implies(x, z)) }

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(or) = {2}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [implies](x1, x2) = [2] x1 + [2] x2 + [5]
                                             
            [not](x1) = [1] x1 + [2]         
                                             
         [or](x1, x2) = [1] x1 + [1] x2 + [2]
  
  The order satisfies the following ordering constraints:
  
         [implies(x, or(y, z))] = [2] x + [2] y + [2] z + [9] 
                                > [2] x + [1] y + [2] z + [7] 
                                = [or(y, implies(x, z))]      
                                                              
           [implies(not(x), y)] = [2] x + [2] y + [9]         
                                > [1] x + [1] y + [2]         
                                = [or(x, y)]                  
                                                              
    [implies(not(x), or(y, z))] = [2] x + [2] y + [2] z + [13]
                                > [2] x + [2] y + [2] z + [9] 
                                = [implies(y, or(x, z))]      
                                                              

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:
  { implies(x, or(y, z)) -> or(y, implies(x, z))
  , implies(not(x), y) -> or(x, y)
  , implies(not(x), or(y, z)) -> implies(y, or(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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