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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if#_A(x1,x2,x3) = max{58, x1 + 56, x3 + 31}
        if_A(x1,x2,x3) = max{x1 + 55, x2 + 19, x3 + 30}
        u_A = 18
        v_A = 29
        true_A = 0
        false_A = 0
  
    precedence:
    
      if# = if = u = v > true = false
    
    partial status:
  
      pi(if#) = []
      pi(if) = []
      pi(u) = []
      pi(v) = []
      pi(true) = []
      pi(false) = []

The next rules are strictly ordered:

  p2

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#(x,if(y,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#(x,if(y,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

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if#_A(x1,x2,x3) = max{x1 + 76, x2 + 28, x3 + 49}
        if_A(x1,x2,x3) = max{x1 + 36, x2 + 13, x3 + 20}
        u_A = 12
        v_A = 19
        true_A = 0
        false_A = 0
  
    precedence:
    
      if# = if = u = v = true = false
    
    partial status:
  
      pi(if#) = [2]
      pi(if) = []
      pi(u) = []
      pi(v) = []
      pi(true) = []
      pi(false) = []

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#(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}


-- Reduction pair.

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

p1: 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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if#_A(x1,x2,x3) = max{x1 + 7, x2 + 3, x3 + 3}
        if_A(x1,x2,x3) = max{x1 + 3, x2 + 3, x3 + 3}
        u_A = 2
        v_A = 1
  
    precedence:
    
      if# = if = u = v
    
    partial status:
  
      pi(if#) = []
      pi(if) = [3]
      pi(u) = []
      pi(v) = []

The next rules are strictly ordered:

  p1

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