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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            purge#_A(x1) = ((0,0),(1,0)) x1 + (3,0)
            ._A(x1,x2) = ((1,0),(1,1)) x1 + (4,3)
            remove_A(x1,x2) = (2,2)
            nil_A() = (1,1)
            if_A(x1,x2,x3) = (0,0)
            =_A(x1,x2) = (1,1)
    
      precedence:
      
        . > purge# = nil = if > remove = =
      
      partial status:
    
        pi(purge#) = []
        pi(.) = [1]
        pi(remove) = []
        pi(nil) = []
        pi(if) = []
        pi(=) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            purge#_A(x1) = (0,0)
            ._A(x1,x2) = x1 + (2,2)
            remove_A(x1,x2) = (3,0)
            nil_A() = (0,1)
            if_A(x1,x2,x3) = (0,0)
            =_A(x1,x2) = (1,1)
    
      precedence:
      
        remove > nil > purge# = . = if = =
      
      partial status:
    
        pi(purge#) = []
        pi(.) = [1]
        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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            remove#_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (2,2)
            ._A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
    
      precedence:
      
        remove# > .
      
      partial status:
    
        pi(remove#) = [1]
        pi(.) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            remove#_A(x1,x2) = x1
            ._A(x1,x2) = x1 + (1,1)
    
      precedence:
      
        remove# = .
      
      partial status:
    
        pi(remove#) = [1]
        pi(.) = [1]
    

The next rules are strictly ordered:

  p1

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