YES

We show the termination of the TRS R:

  ++(nil(),y) -> y
  ++(x,nil()) -> x
  ++(.(x,y),z) -> .(x,++(y,z))
  ++(++(x,y),z) -> ++(x,++(y,z))

-- SCC decomposition.

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

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

and R consists of:

r1: ++(nil(),y) -> y
r2: ++(x,nil()) -> x
r3: ++(.(x,y),z) -> .(x,++(y,z))
r4: ++(++(x,y),z) -> ++(x,++(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: ++(nil(),y) -> y
r2: ++(x,nil()) -> x
r3: ++(.(x,y),z) -> .(x,++(y,z))
r4: ++(++(x,y),z) -> ++(x,++(y,z))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      nil > ++# > . > ++
    
    argument filter:
  
      pi(++#) = [1]
      pi(.) = 2
      pi(++) = [1, 2]
      pi(nil) = []

The next rules are strictly ordered:

  p2, p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: ++(nil(),y) -> y
r2: ++(x,nil()) -> x
r3: ++(.(x,y),z) -> .(x,++(y,z))
r4: ++(++(x,y),z) -> ++(x,++(y,z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: ++(nil(),y) -> y
r2: ++(x,nil()) -> x
r3: ++(.(x,y),z) -> .(x,++(y,z))
r4: ++(++(x,y),z) -> ++(x,++(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      . > ++#
    
    argument filter:
  
      pi(++#) = [1]
      pi(.) = [1, 2]

The next rules are strictly ordered:

  p1

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