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:

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

The next rules are strictly ordered:

  p1

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)
p2: ++#(++(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

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

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: ++#(++(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 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) -> ++#(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:

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

The next rules are strictly ordered:

  p1

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