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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      if#_A(x1,x2,x3) = ((1,1),(1,0)) x1 + ((0,1),(0,0)) x3
      if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + x2 + x3 + (1,1)
      u_A() = (0,0)
      v_A() = (0,0)
      true_A() = (1,1)
      false_A() = (1,1)

The next rules are strictly ordered:

  p1, p2, p3

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