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(x) -> xor(x, true())
  , implies(x, y) -> xor(and(x, y), xor(x, true()))
  , or(x, y) -> xor(and(x, y), xor(x, y))
  , =(x, y) -> xor(x, xor(y, true())) }
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:
  { not(x) -> xor(x, true())
  , implies(x, y) -> xor(and(x, y), xor(x, true()))
  , or(x, y) -> xor(and(x, y), xor(x, y))
  , =(x, y) -> xor(x, xor(y, 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).
  
            [not](x1) = [3] x1 + [3]         
                                             
        [xor](x1, x2) = [1] x1 + [1] x2 + [0]
                                             
               [true] = [0]                  
                                             
    [implies](x1, x2) = [3] x1 + [3] x2 + [3]
                                             
        [and](x1, x2) = [1] x1 + [1] x2 + [2]
                                             
         [or](x1, x2) = [3] x1 + [3] x2 + [3]
                                             
          [=](x1, x2) = [3] x1 + [3] x2 + [3]
  
  This order satisfies the following ordering constraints:
  
           [not(x)] = [3] x + [3]                     
                    > [1] x + [0]                     
                    = [xor(x, true())]                
                                                      
    [implies(x, y)] = [3] x + [3] y + [3]             
                    > [2] x + [1] y + [2]             
                    = [xor(and(x, y), xor(x, true()))]
                                                      
         [or(x, y)] = [3] x + [3] y + [3]             
                    > [2] x + [2] y + [2]             
                    = [xor(and(x, y), xor(x, y))]     
                                                      
          [=(x, y)] = [3] x + [3] y + [3]             
                    > [1] x + [1] y + [0]             
                    = [xor(x, xor(y, true()))]        
                                                      

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(x) -> xor(x, true())
  , implies(x, y) -> xor(and(x, y), xor(x, true()))
  , or(x, y) -> xor(and(x, y), xor(x, y))
  , =(x, y) -> xor(x, xor(y, true())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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