YES(O(1),O(1))

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

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
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(1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(not) = {}, safe(xor) = {1, 2}, safe(true) = {},
   safe(implies) = {2}, safe(and) = {1, 2}, safe(or) = {1, 2},
   safe(=) = {1}
  
  and precedence
  
   not > =, or > =, not ~ or .
  
  Following symbols are considered recursive:
  
   {}
  
  The recursion depth is 0.
  
  For your convenience, here are the satisfied ordering constraints:
  
          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))     
                                                           
          =(y; x) > 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(1))