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:

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          remove#_A(x1,x2) = x2 + (0,1)
          ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      . > remove#
    
    partial status:
  
      pi(remove#) = [2]
      pi(.) = [2]

The next rules are strictly ordered:

  p1

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