YES

We show the termination of the TRS R:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        if# > if
      
      argument filter:
    
        pi(if#) = [1, 2, 3]
        pi(if) = [1, 2, 3]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        if# > if
      
      argument filter:
    
        pi(if#) = [1, 2, 3]
        pi(if) = 2
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        if > if#
      
      argument filter:
    
        pi(if#) = 1
        pi(if) = 2
    

The next rules are strictly ordered:

  p1, p2, p3

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