YES

Problem:
 not(x) -> if(x,false(),true())
 and(x,y) -> if(x,y,false())
 or(x,y) -> if(x,true(),y)
 implies(x,y) -> if(x,y,true())
 =(x,x) -> true()
 =(x,y) -> if(x,y,not(y))
 if(true(),x,y) -> x
 if(false(),x,y) -> y
 if(x,x,if(x,false(),true())) -> true()
 =(x,y) -> if(x,y,if(y,false(),true()))

Proof:
 Bounds Processor:
  bound: 3
  enrichment: roof
  automaton:
   final states: {1}
   transitions:
    not0(1) -> 1*
    if0(1,1,1) -> 1*
    false0() -> 1*
    true0() -> 1*
    and0(1,1) -> 1*
    or0(1,1) -> 1*
    implies0(1,1) -> 1*
    =0(1,1) -> 1*
    if1(1,1,1) -> 1*
    false1() -> 1*
    true1() -> 1*
    not1(1) -> 1*
    true2() -> 1*
    if2(1,1,1) -> 1*
    false2() -> 1*
    true3() -> 1*
  problem:
   
  Qed