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. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = x1 + 4
      if_A(x1,x2,x3) = max{x1 + 2, x2, x3}
    
    2. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = 0
      if_A(x1,x2,x3) = 0
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: if#(if(x,y,z),u,v) -> if#(z,u,v)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = x1
      if_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = 0
      if_A(x1,x2,x3) = 0
    

The next rules are strictly ordered:

  p1, p2

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