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

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(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}


-- 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(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

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        if#_A(x1,x2,x3) = x1 + x2 + x3
        if_A(x1,x2,x3) = x1 + x2 + x3 + 1
        u_A() = 0
        v_A() = 0
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        if#_A(x1,x2,x3) = x1 + x2 + x3
        if_A(x1,x2,x3) = x1 + x2 + x3 + 1
        u_A() = 0
        v_A() = 0
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4

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