YES

We show the termination of the TRS R:

  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(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(if(x,y,z),u,v) -> if(x,if(y,u,v),if(z,u,v))

The set of usable rules consists of

  r1

Take the reduction pair:

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

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(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(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{0, x1 - 2}
        if_A(x1,x2,x3) = max{x1 + 3, x2 + 1, x3}
  
    precedence:
    
      if# = if
    
    partial status:
  
      pi(if#) = []
      pi(if) = [3]

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(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(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{0, x1 - 2}
        if_A(x1,x2,x3) = max{x1 + 3, x2 + 3, x3 + 1}
  
    precedence:
    
      if# = if
    
    partial status:
  
      pi(if#) = []
      pi(if) = [3]

The next rules are strictly ordered:

  p1

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