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. max/plus interpretations on natural numbers:
    
      purge#_A(x1) = x1 + 2
      ._A(x1,x2) = max{x1 + 4, x2 - 1}
      remove_A(x1,x2) = max{x1 + 4, x2 - 2}
      nil_A = 1
      if_A(x1,x2,x3) = max{1, x1 - 4, x2 - 1, x3 - 6}
      =_A(x1,x2) = x2 + 6
    
    2. max/plus interpretations on natural numbers:
    
      purge#_A(x1) = max{0, x1 - 2}
      ._A(x1,x2) = 3
      remove_A(x1,x2) = 1
      nil_A = 0
      if_A(x1,x2,x3) = 0
      =_A(x1,x2) = 0
    

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. max/plus interpretations on natural numbers:
    
      remove#_A(x1,x2) = x2
      ._A(x1,x2) = max{x1 + 1, x2 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      remove#_A(x1,x2) = 0
      ._A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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