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

  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 reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          if#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + (1,3)
          if_A(x1,x2,x3) = x1 + ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (4,2)
          u_A() = (2,1)
          v_A() = (3,8)
  
    precedence:
    
      if# = if = u = v
    
    partial status:
  
      pi(if#) = []
      pi(if) = [1]
      pi(u) = []
      pi(v) = []

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          if#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
          if_A(x1,x2,x3) = ((1,0),(0,0)) x2 + (4,8)
          u_A() = (1,1)
          v_A() = (2,7)
  
    precedence:
    
      if# = if = u = v
    
    partial status:
  
      pi(if#) = []
      pi(if) = []
      pi(u) = []
      pi(v) = []

The next rules are strictly ordered:

  p1

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