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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((1,1),(0,0)) x2
        and_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
        or_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((0,1),(0,0)) x2
        and_A(x1,x2) = x1 + x2 + (1,1)
        or_A(x1,x2) = ((0,1),(0,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = (0,0)
        and_A(x1,x2) = ((1,1),(1,0)) x2 + (1,1)
        or_A(x1,x2) = (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        or#_A(x1,x2) = ((1,0),(1,0)) x2
        or_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        or#_A(x1,x2) = ((1,1),(1,0)) x2
        or_A(x1,x2) = ((1,0),(1,0)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        or#_A(x1,x2) = x2
        or_A(x1,x2) = ((1,1),(1,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

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