YES

We show the termination of the TRS R:

  merge(x,nil()) -> x
  merge(nil(),y) -> y
  merge(++(x,y),++(u(),v())) -> ++(x,merge(y,++(u(),v())))
  merge(++(x,y),++(u(),v())) -> ++(u(),merge(++(x,y),v()))

-- SCC decomposition.

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

p1: merge#(++(x,y),++(u(),v())) -> merge#(y,++(u(),v()))
p2: merge#(++(x,y),++(u(),v())) -> merge#(++(x,y),v())

and R consists of:

r1: merge(x,nil()) -> x
r2: merge(nil(),y) -> y
r3: merge(++(x,y),++(u(),v())) -> ++(x,merge(y,++(u(),v())))
r4: merge(++(x,y),++(u(),v())) -> ++(u(),merge(++(x,y),v()))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: merge#(++(x,y),++(u(),v())) -> merge#(y,++(u(),v()))

and R consists of:

r1: merge(x,nil()) -> x
r2: merge(nil(),y) -> y
r3: merge(++(x,y),++(u(),v())) -> ++(x,merge(y,++(u(),v())))
r4: merge(++(x,y),++(u(),v())) -> ++(u(),merge(++(x,y),v()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      merge#_A(x1,x2) = ((1,0,0),(1,0,0),(0,1,0)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      ++_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (1,1,1)
      u_A() = (0,0,0)
      v_A() = (0,0,0)

The next rules are strictly ordered:

  p1

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