YES

We show the termination of the TRS R:

  purge(nil()) -> nil()
  purge(.(x,y)) -> .(x,purge(remove(x,y)))
  remove(x,nil()) -> nil()
  remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))

-- SCC decomposition.

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

p1: purge#(.(x,y)) -> purge#(remove(x,y))
p2: purge#(.(x,y)) -> remove#(x,y)
p3: remove#(x,.(y,z)) -> remove#(x,z)

and R consists of:

r1: purge(nil()) -> nil()
r2: purge(.(x,y)) -> .(x,purge(remove(x,y)))
r3: remove(x,nil()) -> nil()
r4: remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}


-- Reduction pair.

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

p1: purge#(.(x,y)) -> purge#(remove(x,y))

and R consists of:

r1: purge(nil()) -> nil()
r2: purge(.(x,y)) -> .(x,purge(remove(x,y)))
r3: remove(x,nil()) -> nil()
r4: remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))

The set of usable rules consists of

  r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          purge#_A(x1) = x1
          ._A(x1,x2) = max{x1 + 2, x2}
          remove_A(x1,x2) = max{x1 + 2, x2}
          nil_A = 1
          if_A(x1,x2,x3) = max{x1 + 1, x2, x3}
          =_A(x1,x2) = max{x1 + 1, x2 + 1}
    
      precedence:
      
        nil = = > . > purge# = remove > if
      
      partial status:
    
        pi(purge#) = [1]
        pi(.) = [1, 2]
        pi(remove) = [1, 2]
        pi(nil) = []
        pi(if) = [1, 2]
        pi(=) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          purge#_A(x1) = x1 + 8
          ._A(x1,x2) = max{4, x1 - 9, x2 + 3}
          remove_A(x1,x2) = max{x1 + 2, x2 - 4}
          nil_A = 1
          if_A(x1,x2,x3) = max{x1 - 17, x2 + 4}
          =_A(x1,x2) = max{x1 + 20, x2 + 5}
    
      precedence:
      
        purge# = . = remove = nil = if = =
      
      partial status:
    
        pi(purge#) = []
        pi(.) = [2]
        pi(remove) = []
        pi(nil) = []
        pi(if) = []
        pi(=) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: remove#(x,.(y,z)) -> remove#(x,z)

and R consists of:

r1: purge(nil()) -> nil()
r2: purge(.(x,y)) -> .(x,purge(remove(x,y)))
r3: remove(x,nil()) -> nil()
r4: remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          remove#_A(x1,x2) = max{0, x2 - 2}
          ._A(x1,x2) = max{3, x1, x2 + 1}
    
      precedence:
      
        remove# = .
      
      partial status:
    
        pi(remove#) = []
        pi(.) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          remove#_A(x1,x2) = 0
          ._A(x1,x2) = x2
    
      precedence:
      
        remove# = .
      
      partial status:
    
        pi(remove#) = []
        pi(.) = [2]
    

The next rules are strictly ordered:

  p1

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