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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        purge#_A(x1) = ((1,1),(0,0)) x1
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,0)) x2 + (1,2)
        remove_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,1)
        nil_A() = (1,1)
        if_A(x1,x2,x3) = ((1,0),(1,0)) x1
        =_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        purge#_A(x1) = x1
        ._A(x1,x2) = x1 + (1,1)
        remove_A(x1,x2) = x2 + (2,2)
        nil_A() = (3,1)
        if_A(x1,x2,x3) = ((0,1),(0,0)) x1 + (2,4)
        =_A(x1,x2) = x2 + (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        remove#_A(x1,x2) = x1 + ((1,1),(1,1)) x2
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        remove#_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        ._A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,0)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1

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