YES

We show the termination of the TRS R:

  if(true(),x,y) -> x
  if(false(),x,y) -> y
  if(x,y,y) -> y
  if(if(x,y,z),u,v) -> if(x,if(y,u,v),if(z,u,v))
  if(x,if(x,y,z),z) -> if(x,y,z)
  if(x,y,if(x,y,z)) -> if(x,y,z)

-- SCC decomposition.

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

p1: if#(if(x,y,z),u,v) -> if#(x,if(y,u,v),if(z,u,v))
p2: if#(if(x,y,z),u,v) -> if#(y,u,v)
p3: if#(if(x,y,z),u,v) -> if#(z,u,v)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: if(x,y,y) -> y
r4: if(if(x,y,z),u,v) -> if(x,if(y,u,v),if(z,u,v))
r5: if(x,if(x,y,z),z) -> if(x,y,z)
r6: if(x,y,if(x,y,z)) -> if(x,y,z)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: if#(if(x,y,z),u,v) -> if#(x,if(y,u,v),if(z,u,v))
p2: if#(if(x,y,z),u,v) -> if#(z,u,v)
p3: if#(if(x,y,z),u,v) -> if#(y,u,v)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: if(x,y,y) -> y
r4: if(if(x,y,z),u,v) -> if(x,if(y,u,v),if(z,u,v))
r5: if(x,if(x,y,z),z) -> if(x,y,z)
r6: if(x,y,if(x,y,z)) -> if(x,y,z)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = x1 + 2
          if_A(x1,x2,x3) = max{x1 + 3, x2, x3}
          true_A = 0
          false_A = 0
    
      precedence:
      
        if# = if = true = false
      
      partial status:
    
        pi(if#) = [1]
        pi(if) = [1, 2, 3]
        pi(true) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = x1 + 4
          if_A(x1,x2,x3) = max{x1, x2 + 2, x3 + 2}
          true_A = 0
          false_A = 0
    
      precedence:
      
        if# = if = true = false
      
      partial status:
    
        pi(if#) = [1]
        pi(if) = [1, 2, 3]
        pi(true) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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