YES

We show the termination of the TRS R:

  f(nil()) -> nil()
  f(.(nil(),y)) -> .(nil(),f(y))
  f(.(.(x,y),z)) -> f(.(x,.(y,z)))
  g(nil()) -> nil()
  g(.(x,nil())) -> .(g(x),nil())
  g(.(x,.(y,z))) -> g(.(.(x,y),z))

-- SCC decomposition.

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

p1: f#(.(nil(),y)) -> f#(y)
p2: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))
p3: g#(.(x,nil())) -> g#(x)
p4: g#(.(x,.(y,z))) -> g#(.(.(x,y),z))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p3, p4}


-- Reduction pair.

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

p1: f#(.(nil(),y)) -> f#(y)
p2: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1
      ._A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (1,1,1,1)
      nil_A() = (1,1,1,1)

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: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(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: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,1,0,0)) x1
      ._A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,0,1)) x1 + (1,1,1,1)

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: g#(.(x,nil())) -> g#(x)
p2: g#(.(x,.(y,z))) -> g#(.(.(x,y),z))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      ._A(x1,x2) = x1 + ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,1,0)) x2 + (1,1,1,1)
      nil_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1, p2

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