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()))

-- 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()))

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()))

The set of usable rules consists of

  r1, r2, r3, r4

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) = max{x1 + 39, x2 + 31, x3 + 20}
          if_A(x1,x2,x3) = max{x1 + 17, x2 + 10, x3 + 13}
          u_A = 9
          v_A = 0
          true_A = 0
          false_A = 0
    
      precedence:
      
        if# = if = u = true = false > v
      
      partial status:
    
        pi(if#) = [1, 2, 3]
        pi(if) = [1, 2, 3]
        pi(u) = []
        pi(v) = []
        pi(true) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{x1 + 20, x2 + 6, x3 + 14}
          if_A(x1,x2,x3) = max{x1 + 4, x2 + 5, x3 + 5}
          u_A = 4
          v_A = 2
          true_A = 0
          false_A = 0
    
      precedence:
      
        if# = if = u = v = true = false
      
      partial status:
    
        pi(if#) = []
        pi(if) = [2, 3]
        pi(u) = []
        pi(v) = []
        pi(true) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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