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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if#_A(x1,x2,x3) = x1 + x3
            if_A(x1,x2,x3) = x1 + x2 + x3
            u_A() = (0,0)
            v_A() = (0,0)
            true_A() = (1,1)
            false_A() = (1,1)
    
      precedence:
      
        if# = if = v > u = true = false
      
      partial status:
    
        pi(if#) = [1, 3]
        pi(if) = [1, 2, 3]
        pi(u) = []
        pi(v) = []
        pi(true) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            if#_A(x1,x2,x3) = ((0,0),(1,0)) x3 + (3,0)
            if_A(x1,x2,x3) = x1 + ((1,0),(0,0)) x3 + (4,7)
            u_A() = (2,6)
            v_A() = (1,0)
            true_A() = (1,1)
            false_A() = (1,1)
    
      precedence:
      
        if = u = v > if# = true = false
      
      partial status:
    
        pi(if#) = []
        pi(if) = [1]
        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:

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

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#(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}


-- 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())

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:

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