YES

We show the termination of the TRS R:

  and(x,or(y,z)) -> or(and(x,y),and(x,z))
  and(x,and(y,y)) -> and(x,y)
  or(or(x,y),and(y,z)) -> or(x,y)
  or(x,and(x,y)) -> x
  or(true(),y) -> true()
  or(x,false()) -> x
  or(x,x) -> x
  or(x,or(y,y)) -> or(x,y)
  and(x,true()) -> x
  and(false(),y) -> false()
  and(x,x) -> x

-- SCC decomposition.

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

p1: and#(x,or(y,z)) -> or#(and(x,y),and(x,z))
p2: and#(x,or(y,z)) -> and#(x,y)
p3: and#(x,or(y,z)) -> and#(x,z)
p4: and#(x,and(y,y)) -> and#(x,y)
p5: or#(x,or(y,y)) -> or#(x,y)

and R consists of:

r1: and(x,or(y,z)) -> or(and(x,y),and(x,z))
r2: and(x,and(y,y)) -> and(x,y)
r3: or(or(x,y),and(y,z)) -> or(x,y)
r4: or(x,and(x,y)) -> x
r5: or(true(),y) -> true()
r6: or(x,false()) -> x
r7: or(x,x) -> x
r8: or(x,or(y,y)) -> or(x,y)
r9: and(x,true()) -> x
r10: and(false(),y) -> false()
r11: and(x,x) -> x

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}
  {p5}


-- Reduction pair.

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

p1: and#(x,and(y,y)) -> and#(x,y)
p2: and#(x,or(y,z)) -> and#(x,z)
p3: and#(x,or(y,z)) -> and#(x,y)

and R consists of:

r1: and(x,or(y,z)) -> or(and(x,y),and(x,z))
r2: and(x,and(y,y)) -> and(x,y)
r3: or(or(x,y),and(y,z)) -> or(x,y)
r4: or(x,and(x,y)) -> x
r5: or(true(),y) -> true()
r6: or(x,false()) -> x
r7: or(x,x) -> x
r8: or(x,or(y,y)) -> or(x,y)
r9: and(x,true()) -> x
r10: and(false(),y) -> false()
r11: and(x,x) -> x

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:
      
        and# > or > and
      
      argument filter:
    
        pi(and#) = [1, 2]
        pi(and) = [1, 2]
        pi(or) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        and# > or > and
      
      argument filter:
    
        pi(and#) = 2
        pi(and) = [1, 2]
        pi(or) = [1, 2]
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: or#(x,or(y,y)) -> or#(x,y)

and R consists of:

r1: and(x,or(y,z)) -> or(and(x,y),and(x,z))
r2: and(x,and(y,y)) -> and(x,y)
r3: or(or(x,y),and(y,z)) -> or(x,y)
r4: or(x,and(x,y)) -> x
r5: or(true(),y) -> true()
r6: or(x,false()) -> x
r7: or(x,x) -> x
r8: or(x,or(y,y)) -> or(x,y)
r9: and(x,true()) -> x
r10: and(false(),y) -> false()
r11: and(x,x) -> x

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:
      
        or > or#
      
      argument filter:
    
        pi(or#) = [1, 2]
        pi(or) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        or > or#
      
      argument filter:
    
        pi(or#) = [2]
        pi(or) = [1, 2]
    

The next rules are strictly ordered:

  p1

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