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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        merge#_A(x1,x2) = x1 + x2
        ++_A(x1,x2) = x1 + x2 + 1
        u_A() = 0
        v_A() = 0
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        merge#_A(x1,x2) = x1 + x2
        ++_A(x1,x2) = x1 + x2 + 1
        u_A() = 0
        v_A() = 0
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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