YES

We show the termination of the TRS R:

  not(x) -> xor(x,true())
  or(x,y) -> xor(and(x,y),xor(x,y))
  implies(x,y) -> xor(and(x,y),xor(x,true()))
  and(x,true()) -> x
  and(x,false()) -> false()
  and(x,x) -> x
  xor(x,false()) -> x
  xor(x,x) -> false()
  and(xor(x,y),z) -> xor(and(x,z),and(y,z))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: not#(x) -> xor#(x,true())
p2: or#(x,y) -> xor#(and(x,y),xor(x,y))
p3: or#(x,y) -> and#(x,y)
p4: or#(x,y) -> xor#(x,y)
p5: implies#(x,y) -> xor#(and(x,y),xor(x,true()))
p6: implies#(x,y) -> and#(x,y)
p7: implies#(x,y) -> xor#(x,true())
p8: and#(xor(x,y),z) -> xor#(and(x,z),and(y,z))
p9: and#(xor(x,y),z) -> and#(x,z)
p10: and#(xor(x,y),z) -> and#(y,z)

and R consists of:

r1: not(x) -> xor(x,true())
r2: or(x,y) -> xor(and(x,y),xor(x,y))
r3: implies(x,y) -> xor(and(x,y),xor(x,true()))
r4: and(x,true()) -> x
r5: and(x,false()) -> false()
r6: and(x,x) -> x
r7: xor(x,false()) -> x
r8: xor(x,x) -> false()
r9: and(xor(x,y),z) -> xor(and(x,z),and(y,z))

The estimated dependency graph contains the following SCCs:

  {p9, p10}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: and#(xor(x,y),z) -> and#(y,z)
p2: and#(xor(x,y),z) -> and#(x,z)

and R consists of:

r1: not(x) -> xor(x,true())
r2: or(x,y) -> xor(and(x,y),xor(x,y))
r3: implies(x,y) -> xor(and(x,y),xor(x,true()))
r4: and(x,true()) -> x
r5: and(x,false()) -> false()
r6: and(x,x) -> x
r7: xor(x,false()) -> x
r8: xor(x,x) -> false()
r9: and(xor(x,y),z) -> xor(and(x,z),and(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        xor > and#
      
      argument filter:
    
        pi(and#) = 1
        pi(xor) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        and# > xor
      
      argument filter:
    
        pi(and#) = []
        pi(xor) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        xor > and#
      
      argument filter:
    
        pi(and#) = []
        pi(xor) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.