YES

We show the termination of the TRS R:

  rev(a()) -> a()
  rev(b()) -> b()
  rev(++(x,y)) -> ++(rev(y),rev(x))
  rev(++(x,x)) -> rev(x)

-- SCC decomposition.

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

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

and R consists of:

r1: rev(a()) -> a()
r2: rev(b()) -> b()
r3: rev(++(x,y)) -> ++(rev(y),rev(x))
r4: rev(++(x,x)) -> rev(x)

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: rev#(++(x,y)) -> rev#(y)
p2: rev#(++(x,x)) -> rev#(x)
p3: rev#(++(x,y)) -> rev#(x)

and R consists of:

r1: rev(a()) -> a()
r2: rev(b()) -> b()
r3: rev(++(x,y)) -> ++(rev(y),rev(x))
r4: rev(++(x,x)) -> rev(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3

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