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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        xor_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((1,0),(1,1)) x1 + x2
        xor_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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