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 path order with precedence:
  
    precedence:
    
      false > true > if > if#
    
    argument filter:
  
      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.